Chapter 4.3, Problem 2QQ

To determine

The correct choice from the given options.

Answer to Problem 2QQ

It has been determined that the given function has a relative maximum at $x=\frac{7}{3}$ and a relative minimum at $x=3$ . Thus, option (C) is the correct choice.

Explanation of Solution

Given:

The function, $f\left(x\right)=\left(x-2\right){\left(x-3\right)}^2$ .

Concept used:

The critical points of a function $f\left(x\right)$ are the points $x=c$ such that $f^{\prime }\left(c\right)=0$ .

If $f^{\prime }\left(x\right)<0$ for $x<c$ and $f^{\prime }\left(x\right)>0$ for $x>c$ , then the function $f\left(x\right)$ has a local minima at $x=c$ and vice versa.

The local minima (or maxima) where the function attains the least (or greatest) value is called the absolute minima (or maxima) of the function.

Calculation:

The given function is $f\left(x\right)=\left(x-2\right){\left(x-3\right)}^2$ .

Differentiating,

  $f^{\prime }\left(x\right)={\left(x-3\right)}^2+2\left(x-2\right)\left(x-3\right)$

Simplifying,

  $f^{\prime }\left(x\right)=\left(x-3\right)\left[\left(x-3\right)+2\left(x-2\right)\right]$

On further simplification,

  $f^{\prime }\left(x\right)=\left(x-3\right)\left[x-3+2x-4\right]$

Finally,

  $f^{\prime }\left(x\right)=\left(x-3\right)\left(3x-7\right)$

Equating the first derivative of the function to obtain critical points,

  $\left(x-3\right)\left(3x-7\right)=0$

This implies that the critical points are $x=3$ and $x=\frac{7}{3}$ .

Now, let $x<\frac{7}{3}<3$ .

It follows that, $x-3<0$ and $3x-7<0$ .

Then, $f^{\prime }\left(x\right)=\left(x-3\right)\left(3x-7\right)>0$ .

Similarly, let $\frac{7}{3}<x<3$ .

It follows that, $x-3<0$ and $3x-7>0$ .

Then, $f^{\prime }\left(x\right)=\left(x-3\right)\left(3x-7\right)<0$ .

Finally, let $\frac{7}{3}<3<x$ .

It follows that, $x-3>0$ and $3x-7>0$ .

Then, $f^{\prime }\left(x\right)=\left(x-3\right)\left(3x-7\right)>0$ .

So, it can be seen that $f^{\prime }\left(x\right)>0$ for $x<\frac{7}{3}$ and $f^{\prime }\left(x\right)<0$ for $x>\frac{7}{3}$ .

This implies that the given function has a relative maximum at $x=\frac{7}{3}$ .

It can also be seen that $f^{\prime }\left(x\right)<0$ for $x<3$ and $f^{\prime }\left(x\right)>0$ for $x>3$ .

This implies that the given function has a relative minimum at $x=3$ .

Thus, the given function has a relative maximum at $x=\frac{7}{3}$ and a relative minimum at $x=3$ .

Conclusion:

It has been determined that the given function has a relative maximum at $x=\frac{7}{3}$ and a relative minimum at $x=3$ . Thus, option (C) is the correct choice.