Calculus
Exam Questions + Memo

√ \(f(x)=12\) is a constant function — its graph is a flat horizontal line, so it never rises or falls as \(x\) changes. Since the derivative measures rate of change, a constant's rate of change is always 0, giving \(f'(x)=0\).

To write a term as a power of \(x\), use the negative-exponent law \(\dfrac{1}{x^n}=x^{-n}\): a variable in the denominator moves to the numerator by flipping the sign of its exponent.

√ \(\dfrac{1}{x^4}=x^{-4}\)

√ The power rule says that for \(x^n\), you bring the exponent down to sit in front as a multiplying coefficient, then subtract 1 from the exponent. Here \(n=5\), so \(\dfrac{d}{dx}\left[x^5\right]=5x^{5-1}=5x^4\).

√ Remember what the derivative actually means: \(f'(a)\) is the gradient of the tangent line to the graph at \(x=a\).

√ So if \(f'(a)=0\), that tangent's gradient is 0 — and a line with gradient 0 is horizontal. So the tangent at \(x=a\) is horizontal (its gradient is 0).

√ First principles means going straight back to the definition of the derivative, \(f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), rather than using a shortcut rule. The first step is to build \(f(x+h)\) by replacing every \(x\) in \(f(x)=3x^2-6\) with \((x+h)\), then expand the brackets: \(f(x+h)=3(x+h)^2-6=3x^2+6xh+3h^2-6\).

√ Now subtract the original \(f(x)=3x^2-6\) from this expanded expression. The \(3x^2\) terms cancel each other, and the \(-6\) terms cancel each other too, leaving only the parts that contain \(h\): \(f(x+h)-f(x)=6xh+3h^2=h(6x+3h)\), where \(h\) is then factored out of both remaining terms.

√ Dividing by \(h\) is what actually forms the difference quotient (the average gradient between \(x\) and \(x+h\)) — and because \(h\) was factored out on top in the line above, it cancels directly with the \(h\) on the bottom: \(\dfrac{f(x+h)-f(x)}{h}=6x+3h\).

√ With the troublesome \(h\) already cancelled out of the denominator, it's now safe to let \(h\to0\): the surviving \(3h\) term simply shrinks to \(0\), leaving the derivative \(f'(x)=6x\).

Work from the definition of the derivative from first principles: \(f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\). So the plan is: write out \(f(x+h)-f(x)\) and simplify it as far as possible, divide by \(h\), and only then let \(h\to0\) — never substitute \(h=0\) before the \(h\) in the denominator has been cancelled, or you get the undefined form \(\frac{0}{0}\).

√ \(f(x+h)-f(x)=\dfrac{2}{3(x+h)}-\dfrac{2}{3x}=\dfrac{2x-2(x+h)}{3x(x+h)}=\dfrac{-2h}{3x(x+h)}\) — the two fractions are combined over the common denominator \(3x(x+h)\), and expanding the numerator, \(2x-2(x+h)=2x-2x-2h=-2h\), leaves a single \(h\) factor on top. That's the key thing to check for at this stage: an \(h\) on top is exactly what will let us cancel with the \(h\) we divide by next.

√ Divide by \(h\): \(\dfrac{-2h}{3x(x+h)}\div h=\dfrac{-2}{3x(x+h)}\). Since every term of \(f(x+h)-f(x)\) already carries a factor of \(h\), dividing by \(h\) is simply cancelling that common factor — this converts the difference quotient into an expression that is safe to evaluate at \(h=0\), which is the whole point of doing this algebra first instead of substituting straight away.

√ Let \(h\to0\): \(f'(x)=-\dfrac{2}{3x^2}\). With the \(h\) already cancelled, letting \(h\to0\) is now just a direct substitution: \((x+h)\to x\), so the denominator \(3x(x+h)\to3x\cdot x=3x^2\). This is the derivative (gradient) function of \(f(x)=\dfrac{2}{3x}\) — it tells you the gradient of the original graph at any value of \(x\) (excluding \(x=0\), where \(f\) itself is undefined).

To differentiate \(f(x)=4x^3-7x^2+2x-9\), apply the power rule to each term on its own: for any term \(ax^n\), the derivative is \(n \cdot a\,x^{n-1}\) — multiply the term by its own exponent, then drop that exponent by 1. This works term-by-term because the derivative of a sum is just the sum of the derivatives.

\(4x^3 \to 3\times4\,x^{3-1}=12x^2\); \; -7x^2 \to 2\times(-7)\,x^{2-1}=-14x\); \; 2x \to 1\times2\,x^{1-1}=2\).

The constant \(-9\) has no \(x\) attached, so it graphs as a flat, horizontal line — zero slope everywhere — which is why its derivative is \(0\) and it simply disappears from the answer.

√ Combining the differentiated terms gives \(f'(x)=12x^2-14x+2\).

√ First principles means using the definition \(f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), so the first step is to build \(f(x+h)\): replace every \(x\) in \(f(x)=5x^2+x-3\) with \((x+h)\) and expand. \(f(x+h)=5(x+h)^2+(x+h)-3=5x^2+10xh+5h^2+x+h-3\)

√ Now subtract the original \(f(x)\) — the \(5x^2\), \(x\) and \(-3\) terms cancel out with their matching terms in \(f(x)\), leaving only the pieces that contain \(h\). Factoring \(h\) out of what remains (every surviving term has an \(h\) in it) sets up the division in the next step: \(f(x+h)-f(x)=10xh+5h^2+h=h(10x+5h+1)\)

√ Divide by \(h\) as the definition requires. Since \(h\) is a common factor top and bottom and \(h\neq0\) at this point (it's still just a small nonzero increment, not yet zero), it cancels directly, leaving an expression with no \(h\) in the denominator: \(\dfrac{f(x+h)-f(x)}{h}=10x+5h+1\)

√ Only now — with the division by \(h\) already gone — is it valid to let \(h\to0\): the remaining \(5h\) term simply disappears, giving the derivative \(f'(x)=10x+1\)

√ Before differentiating, rewrite each term as a power of \(x\) so the power rule can be applied directly: \(\sqrt{x}=x^{\frac12}\) and \(\dfrac{1}{x^2}=x^{-2}\), giving \(f(x)=3x^{\frac12}-2x^{-2}\).

√ Now apply the power rule (\(\dfrac{d}{dx}\left[ax^n\right]=anx^{n-1}\): multiply the coefficient by the exponent, then reduce the exponent by 1) to each term separately. First term: \(3\times\frac12=\frac32\), new exponent \(\frac12-1=-\frac12\). Second term: \(-2\times(-2)=4\), new exponent \(-2-1=-3\). This gives \(f'(x)=\dfrac{3}{2}x^{-\frac12}+4x^{-3}\).

√ Finally, rewrite the negative exponents back as fractions — a negative exponent just means "1 over that power", i.e. \(x^{-n}=\dfrac{1}{x^n}\) — so the derivative is left in the same neat surd/fraction form the question was posed in: \(f'(x)=\dfrac{3}{2\sqrt{x}}+\dfrac{4}{x^3}\).

√ First principles always starts from the definition \(f'(x)=\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), so the first step is to find \(f(x+h)\) by replacing every \(x\) in the function with \(x+h\). Since \(f(x)=-5\) is a constant function — it gives \(-5\) no matter what you substitute — replacing \(x\) with \(x+h\) changes nothing: \(f(x+h)=-5\) as well.

√ Substitute into the numerator: \(f(x+h)-f(x)=-5-(-5)=0\). This makes sense — because the function's output never changes, the difference between its value at \(x+h\) and at \(x\) is always \(0\).

√ Divide by \(h\): \(\dfrac{f(x+h)-f(x)}{h}=\dfrac{0}{h}=0\) for every \(h\ne0\). Zero divided by any nonzero number is still zero, so this ratio (the gradient of the secant line) is \(0\) no matter how big or small \(h\) is.

√ Let \(h\to0\): since the expression is already the constant \(0\) for every \(h\ne0\), sending \(h\) to \(0\) can't change that value, so \(f'(x)=0\). This matches the graph — \(f(x)=-5\) is a horizontal line, and a horizontal line has zero gradient everywhere.

√ The graph of \(h(t)=-5t^2+20t+2\) is a downward-opening parabola in \(t\) (since the coefficient of \(t^2\) is negative), so it rises to a single maximum turning point and then falls back down — at that exact instant the ball is neither still rising nor falling, so its velocity is momentarily zero. Velocity is the rate of change of height, so differentiate \(h(t)\) with respect to \(t\): \(v(t)=h'(t)=-10t+20\)

√ Maximum height occurs where the ball is momentarily at rest: \(v(t)=0\). This is the same idea as finding a turning point on any graph — the gradient is zero exactly at the top of the curve, so setting the velocity (the gradient function) to zero pinpoints the moment of maximum height.

√ Solve this equation for \(t\) by isolating the variable: \(-10t+20=0 \Rightarrow \boxed{t=2\text{ s}}\). So the ball reaches its maximum height 2 seconds after being thrown.

√ (a) Set up first principles. The definition of the derivative is \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\), so the first job is to find \(f(x+h)\): replace every \(x\) in \(f(x)=2x^2+4\) with \((x+h)\) and expand. \(f(x+h)=2(x+h)^2+4=2x^2+4xh+2h^2+4\)

√ Subtract \(f(x)\) and factorise out \(h\). Subtracting \(f(x)=2x^2+4\) cancels the \(2x^2\) and \(+4\) terms exactly, leaving only the terms that contain \(h\) — that's expected, since these are the terms about to be divided by \(h\) in the next step. \(f(x+h)-f(x)=4xh+2h^2=h(4x+2h)\)

√ Cancel the common \(h\), then let \(h\to0\). Dividing by \(h\) is allowed here because in the limit process \(h\) is approaching zero but is never actually equal to zero, so this leaves \(4x+2h\). Now every remaining term with an \(h\) in it simply disappears as \(h\to0\), which is what gives the derivative. \(f'(x)=\lim_{h\to0}(4x+2h)=4x\)

√ (b) Use the given gradient to find \(x\). \(f'(x)=4x\) is the gradient of the tangent at any point on the curve, so to find where that gradient equals \(-12\), set \(f'(x)=-12\) and solve for \(x\). \(4x=-12 \Rightarrow x=-3\)

√ Find the matching \(y\)-value. A tangent line touches the curve at one specific point, so its coordinates are needed — substitute \(x=-3\) into the original function \(f(x)\) (not \(f'(x)\)) to get the corresponding \(y\)-value. \(f(-3)=2(9)+4=22\), point \((-3;22)\)

√ Build the tangent's equation. A straight line has the form \(y=mx+c\); the gradient \(m=-12\) is already known from the question, so substitute the point \((-3;22)\) found above to solve for the remaining unknown, \(c\). Tangent: \(y=-12x+c\); \(22=-12(-3)+c \Rightarrow c=-14\)

√ Final answer. Putting \(m=-12\) and \(c=-14\) together gives the equation of the tangent. \(y=-12x-14\)

√ A tangent line needs a point and a gradient, so we find the point first: substitute \(x=1\) into \(f(x)\) itself to get the \(y\)-coordinate. \(f(1)=1-3+2=0\), so the tangent touches the curve at \((1,0)\).

√ Next we need the gradient at that point, which means differentiating \(f(x)\) to get the gradient function: \(f'(x)=3x^2-6x\).

√ Substitute \(x=1\) into \(f'(x)\) (not \(f(x)\) again) to get the actual gradient of the tangent at that point: \(f'(1)=3-6=-3\).

√ Now use the point \((1,0)\) and gradient \(m=-3\) in the point-gradient form \(y-y_1=m(x-x_1)\): \(y-0=-3(x-1)\).

√ Expand and simplify into standard form: \(y=-3x+3\).

(a) Turning points. A turning point is where the graph stops rising or falling for an instant, so its gradient is momentarily \(0\) — that means we need to solve \(f'(x)=0\).

√ \(f'(x)=3x^2-10x+8=(3x-4)(x-2)=0 \Rightarrow x=\frac43\) or \(x=2\) — differentiate \(f\) term by term to get \(f'(x)\), set it equal to zero, then factorise the resulting quadratic to find the two \(x\)-values where the gradient is flat.

√ \(f(\frac43)=\frac4{27}\); \(f(2)=0\) — an \(x\)-value on its own isn't a full point. Substitute each \(x\)-value back into the original function \(f(x)\) (not \(f'(x)\)!) to find its matching \(y\)-value.

√ Turning points: \((\frac43;\frac4{27})\) and \((2;0)\) — pairing each \(x\) with the \(y\)-value just found gives the two coordinates the question actually asks for.

(b) \(f'(x)=3x^2-10x+8\) is itself a quadratic. Its leading coefficient (\(3\)) is positive, so its graph is an upward parabola with roots at the two \(x\)-values found in (a). An upward parabola sits above the \(x\)-axis (positive) outside its roots, and below it (negative) between them.

√ (b) \(f'(x)>0\) for \(x<\frac43\) or \(x>2\) — this also matches the shape of \(f\) itself: these are exactly the intervals where \(f\) is increasing, i.e. everywhere except the dip between its two turning points.

(c) First differentiate \(f'(x)\) once more to get \(f''(x)\), then build the exact expression the question asks about, \(x\cdot f''(x)\).

√ \(f''(x)=6x-10\); \(x\cdot f''(x)=6x^2-10x\), roots at \(x=0\) and \(x=\frac53\) — multiplying \(f''(x)\) by \(x\) gives a brand-new quadratic; setting it to \(0\) and solving finds the boundary points where this new expression could change sign.

√ The product is negative between its roots: \(0<x<\frac53\) — just like in part (b), \(6x^2-10x\) has a positive leading coefficient (\(6\)), so it's an upward parabola that is negative strictly between its two roots — giving the required interval.

√ A point of inflection is where the graph changes concavity — from curving upward (concave up) to curving downward (concave down), or the reverse. This happens exactly where the second derivative equals zero, so we first need \(g'(x)\) in order to then find \(g''(x)\): differentiating \(g(x)=2x^3-9x^2+12x+5\) once gives \(g'(x)=6x^2-18x+12\).

√ Differentiating a second time gives \(g''(x)=12x-18\). Setting \(g''(x)=0\) and solving locates the x-value where the concavity switches: \(12x-18=0 \Rightarrow x=1.5\).

√ This x-value only tells us where the inflection happens — to get the actual point on the graph, substitute \(x=1.5\) back into the original function \(g(x)\), not into \(g'(x)\) or \(g''(x)\): \(g(1.5)=2(3.375)-9(2.25)+18+5=9.5\).

√ Point of inflection: \((1.5;9.5)\) — this is the exact point where the curve of \(g\) stops bending one way and starts bending the other.

(a) x-intercepts

√ A cubic doesn't factorise by inspection like a quadratic does, so start with the Factor Theorem: test values that divide into the constant term (\(-16\)) to find one root. Trying \(x=-2\): \(f(-2)=-8+24-16=0\), so \((x+2)\) is a factor of \(f(x)\).

√ Dividing \(f(x)\) by \((x+2)\) leaves a quadratic, and factorising that quadratic completely factorises the cubic: \(x^3-12x-16=(x+2)(x^2-2x-8)=(x+2)^2(x-4)\).

√ Each factor set equal to zero gives an x-intercept: \(x=-2\) (repeated) or \(x=4\). Because \((x+2)\) is squared, the graph doesn't cross the x-axis at \(x=-2\) — it only touches it and turns back. Keep that in mind, because it reappears as a turning point in part (b).

(b) Turning points

√ A turning point is where the gradient is momentarily zero, so it's found by solving \(f'(x)=0\), not \(f(x)=0\): \(f'(x)=3x^2-12=0 \Rightarrow x=\pm2\).

√ This only gives the x-coordinates of the turning points. To find the matching y-coordinates, substitute each value back into the original function \(f(x)\) (never into \(f'(x)\)): \(f(-2)=0\) — which matches the repeated x-intercept from part (a), confirming that this turning point sits exactly on the x-axis; and \(f(2)=8-24-16=-32\).

√ So the turning points are \((-2;0)\) and \((2;-32)\).

(c) Concavity

√ Concavity is read from the second derivative: the graph is concave up where \(f''(x)>0\) (curving upward, like the inside of a bowl) and concave down where \(f''(x)<0\). Here \(f''(x)=6x\), so solving \(f''(x)>0\) gives \(x>0\) — the graph of \(f\) is concave up for \(x>0\).

√ Step 1 — find the gradient of the given line. A tangent that is parallel to a line must have the exact same gradient as that line, so first rewrite \(6x-y+2=0\) into the familiar \(y=mx+c\) form so the gradient can be read off directly: \(6x-y+2=0 \Rightarrow y=6x+2\), so \(m=6\).

√ Step 2 — find where on the curve \(f\) has this gradient. \(f'(x)\) gives the gradient of the tangent to the curve at any value of \(x\), so to find the point where that gradient equals 6, set \(f'(x)\) equal to 6 and solve for \(x\): \(f'(x)=2x+4=6 \Rightarrow x=1\).

√ Step 3 — find the matching \(y\)-coordinate. \(x=1\) only tells us the position along the curve — to get the actual point where the tangent touches \(f\), substitute \(x=1\) back into the original function \(f(x)\), not \(f'(x)\): \(f(1)=1+4+1=6\), giving the point \((1;6)\).

√ Step 4 — build the equation of the tangent. A straight line is fixed once you know one point on it and its gradient — here that's the point \((1;6)\) from Step 3 and the gradient \(m=6\) matched in Step 1. Substitute both into the point-gradient form \(y-y_1=m(x-x_1)\) and simplify: \(y-6=6(x-1) \Rightarrow y=6x-6+6=6x\). This is the equation of the required tangent.

√ Stationary points are where the graph is momentarily flat, i.e. where the gradient \(f'(x)=0\), so start by differentiating and factorising: \(f'(x)=3x^2-12=3(x-2)(x+2)=0 \Rightarrow x=-2\) or \(x=2\). These are the only two candidates for a local maximum or minimum.

√ \(x=-2\) and \(x=2\) split the number line into three intervals. To find the sign of \(f'(x)\) on each one (without using \(f''\)), pick any convenient test value inside each interval and substitute it into \(f'(x)\) — the sign tells you whether the graph is rising or falling there: \(f'(-3)=3(9)>0\); \(f'(0)=3(-4)<0\); \(f'(3)=3(9)>0\).

√ At \(x=-2\): \(f'\) changes \(+\to-\) — the graph is increasing just before \(x=-2\) and decreasing just after it, i.e. rising then falling, which is exactly the shape of a local maximum. Substitute \(x=-2\) into the original \(f(x)\) (not \(f'(x)\)) to get the actual height of the turning point: \(f(-2)=-8+24=16\), a local maximum at \((-2;16)\).

√ At \(x=2\): \(f'\) changes \(-\to+\) — falling then rising, the mirror-image shape, which is what identifies a local minimum. Substituting \(x=2\) into \(f(x)\) gives \(f(2)=8-24=-16\), a local minimum at \((2;-16)\).

Part (a): showing \(S(r)=2\pi r^2+\dfrac{108\pi}{r}\)

√ The can's volume is fixed at \(54\pi\text{ cm}^3\) — that's the constraint linking \(r\) and \(h\) together, so start there: \(V=\pi r^2h=54\pi\). We can't yet write \(S\) purely in terms of \(r\) while \(h\) is still an unknown of its own, so the first job is to use this equation to express \(h\) in terms of \(r\). Dividing both sides by \(\pi r^2\) gives \(h=\dfrac{54}{r^2}\).

√ Now think about what "total surface area" means for a closed can: it has a circular base, a circular lid, and a curved side wall wrapped around it. The two circles together contribute \(2\pi r^2\), and the curved wall — if you imagine unrolling it flat — becomes a rectangle of width \(2\pi r\) (the circumference) and height \(h\), giving \(2\pi rh\). So \(S(r)=2\pi r^2+2\pi rh\).

√ This still has \(h\) in it, which is exactly why the substitution from the first line matters: replacing \(h\) with \(\dfrac{54}{r^2}\) turns \(S\) into a function of \(r\) alone, which is what the question asks for. \(S(r)=2\pi r^2+2\pi rh=2\pi r^2+2\pi r\left(\dfrac{54}{r^2}\right)=2\pi r^2+\dfrac{108\pi}{r}\), as required.

Part (b): minimising the surface area

√ To find the value of \(r\) that makes \(S(r)\) as small as possible, use the standard optimisation idea: at a minimum (or maximum) turning point, the gradient of the graph is momentarily zero, so we differentiate \(S(r)\) and set \(S'(r)=0\). Differentiating \(2\pi r^2\) gives \(4\pi r\), and differentiating \(108\pi r^{-1}\) gives \(-108\pi r^{-2}\), so \(S'(r)=4\pi r-\dfrac{108\pi}{r^2}=0\). Multiplying through by \(r^2\) (allowed since \(r>0\) — a can can't have a negative or zero radius) clears the fraction: \(4r^3=108\), so \(r^3=27\), giving \(r=3\).

√ Setting \(S'(r)=0\) only tells us \(r=3\) is a stationary point — it could be a minimum, a maximum, or neither, so this needs to be checked before calling it "the" answer. The second derivative test settles it: \(S''(r)=4\pi+\dfrac{216\pi}{r^3}\), and since \(r>0\) means every term here is positive, \(S''(r)>0\) for every \(r>0\). A positive second derivative means the graph is concave up (curving upward) at that point, which confirms \(r=3\) genuinely gives a minimum, not a maximum.

√ With \(r=3\) confirmed as the minimising radius, substitute it back into the original \(S(r)\) formula from part (a) to get the actual minimum surface area: \(S(3)=2\pi(9)+\dfrac{108\pi}{3}=18\pi+36\pi=\boxed{54\pi\text{ cm}^2\approx169.6\text{ cm}^2\text{ at }r=3\text{ cm}}\).

√ (a) \(f'(x)\) is the gradient function of \(f\): at any x-value, the height of the \(f'\) graph tells you the gradient of the tangent to \(f\) at that same x-value. The y-intercept of \(f'\) is the value \(f'(0)\), so it directly gives the gradient of \(f\) at \(x=0\): gradient \(=4\)

√ (b) A turning point of \(f\) is where its tangent is momentarily horizontal, i.e. where the gradient is zero, i.e. where \(f'(x)=0\). On the graph of \(f'\), the points where \(f'(x)=0\) are exactly its x-intercepts. So the x-intercepts of \(f'\) tell you directly where \(f\) turns: \(x=1\) or \(x=3\)

√ (c) \(f\) is decreasing wherever its gradient is negative, i.e. wherever \(f'(x)<0\) — so to find where \(f\) is decreasing, read off the sketch of \(f'\) where its graph lies below the x-axis. Since \(f'\) is the parabola through the two roots found in (b), opening upward as shown, it dips below the x-axis only between those roots. So \(f'(x)<0\), and \(f\) is decreasing, for \(1<x<3\)

√ (a) The wall replaces one side of the rectangle, so the \(200\) m of fencing only has to cover the other three sides: the two sides of length \(x\) (perpendicular to the wall) and the one side running parallel to the wall. That leaves \(200-2x\) metres of fencing for the parallel side, so its length is \(200-2x\). Since area \(=\) length \(\times\) width, \(A(x)=x(200-2x)=200x-2x^2\), which is exactly the expression required.

√ (b) To find the value of \(x\) that gives the maximum area, look for the stationary point of \(A(x)\) — the point where the gradient of the area function is momentarily zero, which is where an increasing area stops increasing and starts decreasing. Differentiating, \(A'(x)=200-4x\). Setting \(A'(x)=0\) gives \(200-4x=0 \Rightarrow x=50\).

√ A stationary point isn't automatically a maximum — it could be a minimum instead — so the second derivative is used to check. \(A''(x)=-4\), and because \(A''(x)<0\) the graph of \(A\) is concave down at \(x=50\), which confirms this stationary point is indeed a maximum.

√ Now that \(x=50\) is confirmed, substitute it back into the expression for the parallel side found in (a): length \(=200-2(50)=200-100=100\) m.

√ Finally, the maximum area is length \(\times\) width using these two dimensions: maximum area \(=50\times100=5000\text{ m}^2\).

√ The gradient of a tangent to the graph of \(p\) at any point is given by the derivative at that point, so first find \(p'(x)\): \(p'(x)=-3x^2-8\)

√ To decide whether a positive gradient is ever possible, check the sign of \(p'(x)\) for every real \(x\): \(x^2\) can never be negative, so multiplying by \(-3\) gives \(-3x^2\leq0\) for all real \(x\), and \(-8<0\), so \(p'(x)\leq-8<0\) for all real \(x\)

√ \(p'(x)\) is always negative — the least negative it can ever be is \(-8\) (when \(x=0\), since \(-3x^2=0\) there), and adding two negative (or zero) quantities can never produce a result of zero or above, so no real value of \(x\) can make the gradient zero or positive

√ No, it is not possible — every tangent to \(p\) has a negative gradient

√ The gradient of the tangent at a point is the derivative evaluated at that point, so first differentiate: \(\dfrac{dy}{dx}=-2x\). At \(x=3\), the gradient is \(-2(3)=-6\).

√ Before using \(P(3;-10)\) in the tangent equation, confirm it actually lies on the curve: \(y(3)=-(3)^2-1=-9-1=-10\) ✓, which matches the given point.

√ Now build the tangent line using the point-gradient form \(y-y_1=m(x-x_1)\), with \(m=-6\) and \((x_1;y_1)=(3;-10)\): \(y-(-10)=-6(x-3)\), which simplifies to \(y=-6x+8\).

√ \(R\) is where the tangent crosses the x-axis, so set \(y=0\): \(0=-6x+8 \Rightarrow x=\frac43\), giving \(R\left(\frac43;0\right)\).

√ \(PT\) is drawn perpendicular to the x-axis with \(T\) on the x-axis, so \(T\) is directly below \(P\): it keeps \(P\)'s x-coordinate but has \(y=0\), giving \(T(3;0)\).

√ Both \(R\) and \(T\) lie on the x-axis, so \(RT\) is simply the horizontal distance between their x-coordinates: \(RT=\left|3-\frac43\right|=\dfrac53\).

(a) x-intercepts

√ To find an x-intercept we need a root of \(f(x)=0\). We search by trial among the factors of the constant term (12), using the Factor Theorem: if \(f(k)=0\) then \((x-k)\) is a factor. Trying \(x=2\): \(f(2)=8-4-16+12=0\), so \(x=2\) is a root and \((x-2)\) is a factor

√ Dividing \(f(x)\) by the known factor \((x-2)\) gives \(x^3-x^2-8x+12=(x-2)(x^2+x-6)\). The quadratic \(x^2+x-6\) factorises further into \((x-2)(x+3)\), so \(x^3-x^2-8x+12=(x-2)(x^2+x-6)=(x-2)^2(x+3)\) — notice \((x-2)\) appears twice

√ Setting \(f(x)=0\) means \((x-2)^2(x+3)=0\), giving x-intercepts: \(x=2\) (repeated) and \(x=-3\). Because \(x=2\) is a repeated (double) root, the graph doesn't cross the x-axis there — it touches the axis and turns back, which is confirmed by the turning point found in part (b) below

(b) Turning points

√ A turning point is where the gradient is momentarily zero, so we need \(f'(x)=0\). Differentiating term by term: \(f'(x)=3x^2-2x-8\). This factorises as \((3x+4)(x-2)\), so \(f'(x)=3x^2-2x-8=(3x+4)(x-2)=0 \Rightarrow x=-\frac43\) or \(x=2\) are the x-values of the two turning points

√ To find the actual coordinates we substitute each x-value back into the original function \(f(x)\) (not \(f'(x)\)) to get its y-value: \(f(-\frac43)=\frac{500}{27}\approx18.52\); \(f(2)=0\)

√ So the turning points are: \((-\frac43;\frac{500}{27})\) and \((2;0)\). Since \(f\) is a cubic with a positive leading coefficient, it rises to the first turning point, falls to the second, then rises again — so \((-\frac43;\frac{500}{27})\) is a local maximum and \((2;0)\) is a local minimum, sitting exactly on the x-axis, matching the repeated root from part (a)

(c) Concavity

√ A graph is concave down wherever its second derivative is negative, i.e. \(f''(x)<0\). Differentiating \(f'(x)=3x^2-2x-8\) again gives \(f''(x)=6x-2\). Solving the inequality: \(f''(x)=6x-2<0 \Rightarrow 6x<2 \Rightarrow x<\frac13\). So \(f\) is concave down for all \(x<\frac13\) (and concave up for \(x>\frac13\), with \(x=\frac13\) the point of inflection where concavity switches)

√ To find x-intercepts of a cubic, first find one root by trial using the Factor Theorem — testing values that divide the constant term (4). Trying \(x=1\): \(f(1)=1-1-4+4=0\), confirming \((x-1)\) is a factor.

√ Dividing \(f(x)\) by \((x-1)\) leaves a quadratic that factorises further, since \(x^2-4\) is a difference of squares: \(x^3-x^2-4x+4=(x-1)(x^2-4)=(x-1)(x-2)(x+2)\).

√ Setting each factor to zero gives the x-intercepts: \(\boxed{x=1,\ x=2,\ x=-2}\) — three distinct real roots, unlike the repeated-root cubics elsewhere on this page, since here no factor is squared.

√ Since \(x=2\) is a repeated (double) root and \(x=-3\) is a single root, the cubic must have the form \(g(x)=a(x-2)^2(x+3)\) for some constant \(a\) still to be found — writing it this way builds both root conditions in automatically.

√ Use the extra piece of given information, \(g(0)=-12\), to solve for \(a\): substituting \(x=0\) gives \(g(0)=a(4)(3)=12a=-12 \Rightarrow a=-1\).

√ With \(a=-1\) known, expand the bracketed form back into standard polynomial form — multiply out \((x-2)(x-2)(x+3)\) first, then distribute the negative sign: \(g(x)=-(x-2)^2(x+3)=-(x^2-4x+4)(x+3)=-(x^3-x^2-8x+12)\).

√ Distributing that final negative sign across every term gives \(\boxed{g(x)=-x^3+x^2+8x-12}\).

√ (a) The graph of \(f'\) tells you the sign of \(f\)'s gradient at every \(x\)-value. At \(x=-1\), the \(f'\) graph crosses from below the x-axis to above it (\(-\to+\)), meaning \(f\)'s gradient goes from negative (falling) to positive (rising) — that pattern is exactly what defines a local minimum, so \(f\) has a \(\boxed{\text{local minimum}}\) at \(x=-1\).

√ (b) At \(x=3\), \(f'\) crosses the opposite way, from above the axis to below it (\(+\to-\)), so \(f\)'s gradient goes from positive (rising) to negative (falling) — the signature of a local maximum. So \(f\) has a \(\boxed{\text{local maximum}}\) at \(x=3\).

√ (c) A point of inflection of \(f\) is where \(f''(x)=0\), and \(f''(x)\) is itself the derivative of \(f'(x)\) — so \(f''=0\) is exactly where the \(f'\) graph has its own turning point (its own gradient is momentarily zero). Reading the vertex of the sketched \(f'\) parabola gives the point of inflection of \(f\) at \(\boxed{x=1}\).

√ To decide speeding up vs. slowing down, we need both velocity and acceleration at the given instant. Differentiate \(s(t)\) once for velocity and once more for acceleration: \(v(t)=s'(t)=6t^2-30t+24=6(t-1)(t-4)\); \(a(t)=v'(t)=12t-30\).

√ Substitute \(t=3\) into the velocity function: \(v(3)=6(2)(-1)=-12\) — negative, so at this instant the particle is moving in the negative direction.

√ Substitute \(t=3\) into the acceleration function: \(a(3)=12(3)-30=6\) — positive, meaning the acceleration is acting in the positive direction, opposite to the direction of motion just found.

√ \(v(3)\) and \(a(3)\) have opposite signs (\(-\) and \(+\)). Acceleration acting against the direction of travel is exactly what slows an object down, so \(\boxed{\text{the particle is slowing down at }t=3}\).