Porisma

Worked solution

Read stationary points and concavity from a gradient graph

The diagram shows y=f′(x)y=f^{\prime}(x). The curve cuts the x-axis at (a,0)(a,0) and (−a,0)(-a,0), has a stationary point at (0,b)(0,b), and has asymptote y=ky=k.

(i) State the x-coordinates of the stationary points of y=f(x)y=f(x) and determine the nature of each.

(ii) State the ranges of xx for which y=f(x)y=f(x) is (a) strictly increasing, (b) concave downwards.

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Given gradient graph f′: horizontal asymptote y=k above the x-axis, zeros at −a and a, and minimum (0,b) below the x-axis. The graph decreases for x<0 and increases for x>0, with positive tails and negative values between the zeros.
Given gradient graph f′: horizontal asymptote y=k above the x-axis, zeros at −a and a, and minimum (0,b) below the x-axis. The graph decreases for x<0 and increases for x>0, with positive tails and negative values between the zeros.
Given gradient graph f′: horizontal asymptote y=k above the x-axis, zeros at −a and a, and minimum (0,b) below the x-axis. The graph decreases for x<0 and increases for x>0, with positive tails and negative values between the zeros.

[2008/IJC/Promo/6]

CJC · Discussion · Question 5

Answer

Maximum at x=−ax=-a, minimum at x=ax=a; strictly increasing for x<−ax<-a or x>ax>a; concave downwards for x<0x<0.

(i) Locate and classify the stationary points

Stationary points of ff occur where f′=0f^{\prime}=0, so x=−a,ax=-a,a.

LocationSign of f′ just beforeSign at the pointSign just afterNature of f
−a+0−Maximum
a−0+Minimum

The source sign charts show rising–horizontal–falling at −a-a and falling–horizontal–rising at aa.

How did this part go?

(ii)(a) Read positive gradients

f′(x)>0f^{\prime}(x)>0 for x<−ax<-a or x>ax>a. These are the printed open ranges of strict increase.

How did this part go?

(ii)(b) Read decreasing gradients

Concavity downwards requires f′′(x)<0f^{\prime\prime}(x)<0, meaning that the graph of f′f^{\prime} decreases. This occurs for x<0x<0, not simply wherever f′f^{\prime} is negative.

How did this part go?

Common mistakes

A negative derivative means decreasing; it does not alone mean concave downwards.

Read the hint