Chapter 4.2, Problem 22E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 22E

The local minimum points are $\left(-\sqrt{5},-16\right),\left(\sqrt{5},-16\right)$ and the local maximum point is $\left(0,9\right)$ .

Explanation of Solution

Given:

  $y=x^4-10x^2+9$ .

Calculation:

Consider the given function,

  $y=x^4-10x^2+9$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}\frac{dy}{dx}=4x^3-10\left(2x\right)+0\\ =4x^3-20x\end{array}$

Now to find the critical values, solve the first derivative by equating it to zero. That is,

  $\begin{array}{c}\frac{dy}{dx}=0\\ 4x^3-20x=0\\ 4x\left(x^2-5\right)=0\\ x=0\text{or}{x}^2-5=0\\ x=0\text{or}x=\pm \sqrt{5}\end{array}$

Therefore $x=-\sqrt{5},0,\sqrt{5}$ are the critical values.

The graph of $f\left(x\right)=x^4-10x^2+9$ falls to the left of $x=-\sqrt{5},\sqrt{5}$ and rises to the right of $x=-\sqrt{5},\sqrt{5}$ .

Therefore $x=-\sqrt{5},\sqrt{5}$ are the points of local minima.

The graph of $f\left(x\right)=x^4-10x^2+9$ rises to the left of $x=0$ and falls to the right of $x=0$ .

Therefore $x=0$ is the point of local maxima.

Now the local minimum values are,

  $\begin{array}{c}f\left(-\sqrt{5}\right)={\left(-\sqrt{5}\right)}^4-10{\left(-\sqrt{5}\right)}^2+9\\ =25-50+9\\ =-16\end{array}$

And,

  $\begin{array}{c}f\left(\sqrt{5}\right)={\left(\sqrt{5}\right)}^4-10{\left(\sqrt{5}\right)}^2+9\\ =25-50+9\\ =-16\end{array}$

And the local maximum value is,

  $\begin{array}{c}f\left(0\right)={\left(0\right)}^4-10{\left(0\right)}^2+9\\ =9\end{array}$

b.

To determine

To find: the intervals in which the given function is increasing.

Answer to Problem 22E

The given function increases in the interval $\left(-\sqrt{5},0\right),\left(\sqrt{5},\infty \right)$ .

Explanation of Solution

Given:

  $y=x^4-10x^2+9$ .

Concept used:

The function increases in the interval at which the first derivative is positive.

Calculation:

Consider the given function,

  $y=x^4-10x^2+9$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}\frac{dy}{dx}=4x^3-10\left(2x\right)+0\\ =4x^3-20x\end{array}$

Now the function increases in the interval for which,

  $\begin{array}{c}\frac{dy}{dx}>0\\ 4x^3-20x>0\\ 4x\left(x^2-5\right)>0\\ \left(x>0\&x^2-5>0\right)\text{or}\left(x<0\&x^2-5<0\right)\\ \left(x>0\&x^2>5\right)\text{or}\left(x<0\&x^2<5\right)\\ \left[x>0\&\left(x<-\sqrt{5}\text{or}x>\sqrt{5}\right)\right]\text{or}\left(x<0\&-\sqrt{5}<x<\sqrt{5}\right)\\ \left(x>0\&x<-\sqrt{5}\right)\text{or}\left(x>0\&x>\sqrt{5}\right)\text{or}\left(-\sqrt{5}<x<0\right)\\ x>\sqrt{5}\text{or}-\sqrt{5}<x<0\end{array}$

Hence the given function increases in the intervals $\left(-\sqrt{5},0\right),\left(\sqrt{5},\infty \right)$ .

c.

To determine

To find: the intervals in which the given function is decreasing.

Answer to Problem 22E

The given function decreases in the intervals $\left(-\infty ,-\sqrt{5}\right),\left(0,\sqrt{5}\right)$ .

Explanation of Solution

Given:

  $y=x^4-10x^2+9$ .

Concept used:

The function decreases in the interval at which the first derivative is negative.

Calculation:

Consider the given function,

  $y=x^4-10x^2+9$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}\frac{dy}{dx}=4x^3-10\left(2x\right)+0\\ =4x^3-20x\end{array}$

Now the function decreases in the interval for which,

  $\begin{array}{c}\frac{dy}{dx}<0\\ 4x^3-20x<0\\ 4x\left(x^2-5\right)<0\\ \left(x>0\&x^2-5<0\right)\text{or}\left(x<0\&x^2-5>0\right)\\ \left(x>0\&x^2<5\right)\text{or}\left(x<0\&x^2>5\right)\\ \left(x>0\&-\sqrt{5}<x<\sqrt{5}\right)\text{or}\left[x<0\&\left(x<-\sqrt{5}\text{or}x>\sqrt{5}\right)\right]\\ 0<x<\sqrt{5}\text{or}\left(x<0\&x<-\sqrt{5}\right)\\ 0<x<\sqrt{5}\text{or}x<-\sqrt{5}\end{array}$

Hence the given function decreases in the intervals $\left(-\infty ,-\sqrt{5}\right),\left(0,\sqrt{5}\right)$ .