## Chapter 12

## Q. 12.1

Determine the position and nature of all maximum and minimum points of the following functions:

(a) y = x²

(b) y = −t² + t + 1

(c) y=\frac{x^3}{3}+\frac{x^2}{2}-2 x+1

(d) y = |t|

## Step-by-Step

## Verified Solution

(a) If y = x², then by differentiation \frac{d y}{d x}=2 x.

Recall that at maximum and minimum points either

(i) \frac{d y}{d x} does not exist, or

(ii) \frac{d y}{d x}=0. We must check both of these conditions.

The function 2x exists for all values of x, and so we move to examine any points

where \frac{\mathrm{dy}}{\mathrm{d} x}=0. So, we have

\frac{d y}{d x}=2 x=0

The equation 2x = 0 has one solution, x = 0.We conclude that a turning point exists at x = 0. Furthermore, from the given function y = x², we see that when x = 0 the value of y is also 0, so a turning point exists at the point with coordinates (0, 0). To determine whether this point is a maximum or minimum we use the first-derivative test and examine the sign of \frac{\mathrm{dy}}{\mathrm{d} x} on either side of x = 0. To the left of x = 0, x is clearly negative and so 2x is also negative. To the right of x = 0, x is positive and so 2x is also positive. Hence y has a minimum at x = 0. A graph of y = x² showing this minimum is given in Figure 12.2.

(b) If y = −t² + t + 1, then y′ = −2t + 1 and this function exists for all values of t. Solving y′ = 0 we have

-2 t+1=0 \quad \text { and so } t=\frac{1}{2}

We conclude that there is a turning point at t=\frac{1}{2}. The y coordinate here is -\left(\frac{1}{2}\right)^2+\frac{1}{2}+1=1 \frac{1}{4}. We now inspect the sign of y′ to the left and to the right of t=\frac{1}{2} . A little to the left, say at t = 0, we see that y′ = −2(0) + 1 = 1 which is positive. A little to the right, say at t = 1, we see that y′ = −2(1) + 1 = −1 which is negative. Hence there is a maximum at the point \left(\frac{1}{2}, 1 \frac{1}{4}\right).

A graph of the function is shown in Figure 12.3.

(c) If y=\frac{x^3}{3}+\frac{x^2}{2}-2 x+1, \text { then } y^{\prime}=x^2+x-2 and this function exists for all values of x. Solving y′ = 0 we find

\begin{aligned}x^2+x-2 & =0 \\(x-1)(x+2) & =0 \\x & =1,-2\end{aligned}

There are therefore two turning points, one at x = 1 and one at x = −2 . We consider each in turn.

At x = 1, we examine the sign of y′ to the left and to the right of x = 1. A little way to the left, say at x = 0, we see that y′ = −2 which is negative. A little to the right, say at x = 2, we see that y′ = 2² + 2 − 2 = 4 which is positive. So the point where x = 1 is a minimum.

At x = −2, we examine the sign of y′ to the left and to the right of x = −2. A little way to the left, say at x = −3, we see that y′ = (−3)² + (−3) − 2 = 4 which is positive. A little to the right, say at x = −1, we see that y′ = (−1)²+(−1)−2 = −2 which is negative. So the point where x = −2 is a maximum.

A graph of the function is shown in Figure 12.4.

(d) Recall that the modulus function y = |t| is defined as follows:

y=|t|=\left\{\begin{array}{cc}-t & t \leqslant 0 \\t & t>0\end{array}\right.

A graph of this function was given in Figure 10.13(a) and this should be looked at before continuing. Note that \frac{\mathrm{d} y}{\mathrm{~d} t}=-1 for t negative, and \frac{\mathrm{d} y}{\mathrm{~d} t}=+1 for t positive. The derivative is not defined at t = 0 because of the corner there. There are no points when \frac{\mathrm{d} y}{\mathrm{~d} t}=0 . Because the derivative is not defined at t = 0 this point requires further scrutiny. To the left of t=0, \frac{\mathrm{d} y}{\mathrm{~d} t}<0 ; to the right, \frac{\mathrm{dy}}{\mathrm{d} t}>0 and so t = 0 is a minimum point.