Chapter 4.2, Problem 20E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 20E

The given function has no local extrema.

Explanation of Solution

Given:

  $f\left(x\right)=e^{-0.5x}$ .

Calculation:

Consider the given function,

  $f\left(x\right)=e^{-0.5x}$

Differentiating with respect to $x$ , it follows that

  $f^{\prime }\left(x\right)=-0.5e^{-0.5x}$

Here, the derivative cannot be zero for any real value of $x$ .

So there are no critical values for the given function.

Hence the given function has no local extrema.

b.

To determine

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

Answer to Problem 20E

There is no interval in which the given function is increasing.

Explanation of Solution

Given:

  $f\left(x\right)=e^{-0.5x}$ .

Concept used:

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

Calculation:

Consider the given function,

  $f\left(x\right)=e^{-0.5x}$

Differentiating with respect to $x$ , it follows that

  $f^{\prime }\left(x\right)=-0.5e^{-0.5x}$

Now the function increases in the interval for which,

  $\begin{array}{c}{f}^{\prime }\left(x\right)>0\\ -0.5e^{-0.5x}>0\\ e^{-0.5x}<0\end{array}$

It is not possible for any real value of $x$ .

Hence there is no interval in which the given function is increasing.

c.

To determine

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

Answer to Problem 20E

The given function is decreasing in the interval $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=e^{-0.5x}$ .

Concept used:

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

Calculation:

Consider the given function,

  $f\left(x\right)=e^{-0.5x}$

Differentiating with respect to $x$ , it follows that

  $f^{\prime }\left(x\right)=-0.5e^{-0.5x}$

Now the function decreases in the interval for which,

  $\begin{array}{c}{f}^{\prime }\left(x\right)<0\\ -0.5e^{-0.5x}<0\\ e^{-0.5x}>0\end{array}$

It is true for all the real values of $x$ .

Hence the given function is decreasing in the interval $\left(-\infty ,\infty \right)$ .