Understanding Engineering Mathematics

Solution to review question 10.1.3

(i) For the functionf(x)= 16 x− 3 x^3 we havef′(x)= 16 − 9 x^2. Solving

f′(x)= 16 − 9 x^2 = 0

gives stationary values atx=±^43. To classify these consider the second

derivativef′′(x)=− 18 x.So,forx=^43 ,

f′′(^43 )=− 24 < 0

and we have amaximum.

Forx=−^43 we have

f′′(−^43 )= 24 > 0

and we have aminimum.

We also note thatf′′(x)=0atx=0, so there is a possibility of a

point of inflection at this point. To investigate this we have to consider

how the gradient is changing on either side ofx=0. This is indicated

by the sign off′′(x):

x< 0 f′′(x)=− 18 x> 0

x> 0 f′′(x)=− 18 x< 0

So forx<0,f′′(x)=(f′(x))′is positive and therefore the gradient

f′(x)is increasing asxincreases and the curve is concave upwards. For

x>0,f′′(x)=(f′(x))′is negative, so the gradientf′(x)is decreasing

andxincreases and the curve is concave downwards. Thusx=0is

a point where the curve changes from concave up to concave down

asxincreases – it is a point of inflection. The curve is sketched in

Figure 10.7.

Note that the complete story about the curve requires quite a careful

study of its derivatives.

(ii) Forf(x)=x+

1

x

we havef′(x)= 1 −

1

x^2

. Solving

f′(x)= 1 −

1

x^2

= 0

gives us stationary values atx=±1. We find

f′′(x)=

2

x^3

and so

f′′( 1 )= 2 > 0

giving us aminimumatx=1, while

f′′(− 1 )=− 2 < 0

giving us amaximumatx=−1.