Second derivatives and concavity

(a) Differentiate twice.

$$\frac{dy}{dx}=3x^{2}-6x,\qquad \frac{d^{2}y}{dx^{2}}=6x-6$$

(b) Substitute: $6(-1)-6=-12$.

(c) The rate of change of the first derivative is the second derivative, so set it to zero: $6x-6=0$ gives $x=1$. Do not stop there. The question wants the $y$-coordinate too, so substitute into the original function: $y=1^{3}-3(1)^{2}=-2$.

There are two things to show. First, $f''(3)=2(3)-6=0$. On its own, that only makes $x=3$ a possible point of inflection.

Second, show the concavity changes. Pick an $x$-value either side and record the sign of $f''(x)$ in a table:

The sign of $f''(x)$ changes from negative to positive, so the curve goes from concave down to concave up.

Concave up means $f''(x)>0$. The denominator $1+x^{2}$ is always positive, so the sign comes from the numerator $2x$, which is positive when $x>0$.

At $x=0$, $f''(0)=0$. That is not concave up, which is why (D) is wrong.

Differentiate with the chain rule to get the second derivative:

$$f''(x)=3x^{2}\cos\left(x^{3}\right)$$

Graph it on your calculator over the domain and count where it crosses the axis. Do not count where it just touches. $\cos\left(x^{3}\right)=0$ when $x^{3}=\pm\frac{\pi}{2}$ or $\pm\frac{3\pi}{2}$, so at $x\approx\pm1.16$ and $x\approx\pm1.68$. Those four are sign changes.

$f''(0)=0$ as well, but $3x^{2}$ never goes negative and $\cos\left(x^{3}\right)$ is positive near $0$, so the concavity does not change there. Counting it gives you (D).

Concavity changes where $y''$ changes sign. Differentiate twice:

$$y'=3x^{2}+12x+7+2\sin x,\qquad y''=6x+12+2\cos x$$

Solve $y''=0$ on your calculator to get $x\approx-1.89$. Check the sign change: $y''(-2)\approx-0.83$ and $y''(-1.8)\approx0.75$. If you solved $y'=0$ instead, you found a stationary point. That is not a change of concavity.

Decreasing means $f'(x)<0$. Since $e^{x}>0$ for every $x$, that needs $x-4<0$, so $x<4$.

Concave up means $f''(x)>0$. Use the product rule on $f'(x)$:

$$f''(x)=e^{x}(x-4)+e^{x}=e^{x}(x-3)$$

That is positive when $x>3$. You need both at once, so take the overlap.

Product rule: $f'(x)=xe^{x}+e^{x}=e^{x}(x+1)$, so $f'(-1)=0$. That rules out (C) and (D).

Differentiate again: $f''(x)=e^{x}(x+1)+e^{x}=e^{x}(x+2)$, so $f''(-1)=e^{-1}>0$. The point is a local minimum.

(a) Substitute into the derivative: $f'(1)=2\ln 1+(\ln 1)^{2}=0+0=0$. The gradient is zero, so there is a stationary point at $x=1$.

(b) Set $f'(x)=0$ and factorise:

$$\ln x\,(2+\ln x)=0\ \Rightarrow\ \ln x=0\ \text{ or }\ \ln x=-2$$

$\ln x=0$ is B. So A is at $x=e^{-2}$, and $f\left(e^{-2}\right)=e^{-2}(-2)^{2}=4e^{-2}$.

(c) Differentiate $f'(x)$ again:

$$f''(x)=\frac{2}{x}+\frac{2\ln x}{x}=\frac{2}{x}\left(1+\ln x\right)$$

Since $x>0$, $f''(x)=0$ only when $\ln x=-1$, so $x=e^{-1}$. For $x<e^{-1}$, $1+\ln x<0$ (concave down), and for $x>e^{-1}$ it is positive (concave up), so the concavity does change.

Differentiate (quotient rule on the second term), then set the derivative to zero:

$$f'(x)=\frac{1}{kx}-\frac{k}{(x+1)^{2}}=0\ \Rightarrow\ (x+1)^{2}=k^{2}x\ \Rightarrow\ x^{2}+(2-k^{2})x+1=0$$

One horizontal point of inflection means this quadratic has a repeated root, so its discriminant is zero: $(2-k^{2})^{2}-4=0$, giving $k^{2}=0$ or $k^{2}=4$. $k=0$ is not allowed, because $f$ divides by $k$.

With $k^{2}=4$ the quadratic is $(x-1)^{2}=0$, so $x=1$. Check it is a point of inflection: $f''(x)=-\frac{1}{kx^{2}}+\frac{2k}{(x+1)^{3}}$ gives $f''(1)=0$ for both $k=2$ and $k=-2$. Because the root is repeated, $f'(x)$ does not change sign at $x=1$, so it is not a turning point.

The full walkthrough of 2020 Paper 1, Question 19 goes through it mark by mark.

Count the rungs. $f$ is a cubic, so $f'$ is a quadratic and $f''$ is a straight line. That rules out (A) and (B). In fact, (A) is $f'$.

The graph of $f$ is concave down on the left and concave up on the right, so $f''$ goes from negative to positive. The line must have a positive gradient.

The point of inflection sits halfway between the two turning points, at about $x=1$, and that is where (C) crosses the axis.

$f'(x)=0$ means a stationary point. The curve is still falling at P, so P is out.

$f''(x)<0$ means concave down, so you want a local maximum. Q and S are both local minimums. R is the only maximum.

This is the graph of $f'$, one rung below $f$. Its $x$-intercepts are the stationary points of $f$, and there are two.

Just after $a$, $f'$ goes from positive to negative, so $f$ goes from increasing to decreasing: a local maximum. At $x=0$, $f'$ goes from negative to positive: a local minimum.

The turning point of $f'$ is not a stationary point of $f$. It is where $f''=0$, so it is a point of inflection of $f$.

Go up one rung. $f''$ is the gradient of $f'$. It is positive for $x<-1$, negative for $-1<x<1$ and positive again for $x>1$.

So $f'$ increases, then decreases, then increases, with a maximum at $x=-1$ and a minimum at $x=1$. That is (D).

(C) has its turning points the wrong way round. (A) and (B) are made of straight lines, so their gradient is constant on each side, which cannot be a parabola.

Start with what you can see on $f$. It has local minimums at $x=0$, $3$ and $6$, and local maximums at $x=1.5$ and $4.5$. The points of inflection sit halfway between, at $x=0.75,\ 2.25,\ 3.75,\ 5.25$.

Diagram 2 is zero at the turning points of $f$ and positive where $f$ is increasing (for $0<x<1.5$), so Diagram 2 is $f'$. Diagram 3 is the same shape flipped, so its signs are wrong for both $f'$ and $f''$.

Diagram 1 is zero at the points of inflection. It is positive at $x=0$, $3$ and $6$, where $f$ is concave up, and negative at the maximums, where $f$ is concave down. It is also zero wherever Diagram 2 turns, which is what you would expect of the gradient of $f'$.

First, decide which graph is which. The dashed graph is zero exactly where the solid graph turns, at $x=\frac{\pi}{3}$, $\pi$ and $\frac{5\pi}{3}$. So the dashed graph is the gradient of the solid one: solid is $f'$ and dashed is $f''$.

The stationary points of $f$ are where $f'=0$: $x=0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$ and $2\pi$. Check $f''$ at each. It is negative at $x=0$ and $x=\frac{4\pi}{3}$ (maximums), and positive at $x=\frac{2\pi}{3}$ and $x=2\pi$ (minimums).

The points of inflection are where $f''=0$ and changes sign: $x=\frac{\pi}{3}$, $\pi$ and $\frac{5\pi}{3}$. Join all of these with a smooth curve. Any vertical position is fine, because you only know the derivatives. One possibility is $f(x)=2\cos\left(\frac{3x}{2}\right)$, drawn in blue.