Increasing and concave down from a derivative sketch · source reference

Printed question

The diagram below shows the graph of $y=f'(x)$. The curve passes through the origin and has turning points at $(-3,0)$ and $(-1.5,-1)$. The $x$-axis and $x=1$ are the two asymptotes of the curve.

(i) Find the range of values of $x$ for which the graph of $y=f(x)$ is strictly increasing and concave downwards. [1]

(ii) State the $x$-coordinates of all the stationary points of the graph of $y=f(x)$ and determine the nature of each point. [2]

[2009/HCI/Prelim/I/3]

Printed working

(i) $y=f(x)$ concave downwards ⇒ $f''(x)<0$, so the gradient of $y=f'(x)$ is negative. $y=f(x)$ strictly increasing ⇒ $f'(x)>0$, so the curve $y=f'(x)$ is above the $x$-axis. Thus $x>1$.

(ii) Stationary points: $f'(x)=0\Rightarrow x=-3,0$.

$x$	$0^-$	$0$	$0^+$
$f'(x)$	negative (falling $f$)	$0$ (horizontal)	positive (rising $f$)

Minimum point at $x=0$.

$x$	$(-3)^-$	$-3$	$(-3)^+$
$f'(x)$	negative (falling $f$)	$0$ (horizontal)	negative (falling $f$)

Stationary point of inflexion at $x=-3$. The drawn gradient has a local maximum there, so its slope changes from positive to negative.

Answer key

(i) $x>1$. (ii) Stationary points $x=-3,0$: minimum at $0$ and stationary point of inflexion at $-3$.

Native diagrams