Chapter 5.2, Problem 21E

a.

To determine

To find the local extrema.

Answer to Problem 21E

The local maximum at $\left(-2,4\right)$

Explanation of Solution

Given information:

The given function is $y=4-\sqrt{x+2}$

Concept used:

Extreme values occur only at critical points and end points.

A point in the interior of the domain of a function f at which $f^{\prime }=0$ or $f^{\prime }$ does not exist is called a critical point of f .

Calculation :

The domain of the function is $\left[-2,\infty \right)$

Take derivative of the function

  $y^{\prime }=-\frac{1}{2\sqrt{x+2}}$

At critical point $x=-2$

  $y^{\prime }$ is undefined and the value of y at this critical point

  $\begin{array}{l}y=4-\sqrt{-2+2}\\ y=4\end{array}$

Therefore,

The local maximum at $\left(-2,4\right)$

b.

To determine

To find the intervals on which the function is increasing.

Answer to Problem 21E

The function increases at none.

Explanation of Solution

Given information:

The given function is $y=4-\sqrt{x+2}$

Concept used:

If $f^{\prime }>0$ at each point of $\left(a,b\right)$ , then f increases on $\left[a,b\right]$

It is assumed that f be continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ .

Calculation :

For increasing function $y^{\prime }$ must be greater than 0.

  $-\frac{1}{2\sqrt{x+2}}>0$

For each value of x inside the domain $\left[-2,\infty \right)$ , $\sqrt{x+2}$ is positive and it is multiplied by a negative sign therefore $y^{\prime }$ is never positive.

Therefore,

The function increases at none.

c.

To determine

To find the intervals on which the function is decreasing.

Answer to Problem 21E

The function decreases on $\left[-2,\infty \right)$

Explanation of Solution

Given information:

The given function is $y=4-\sqrt{x+2}$

Concept used:

If $f^{\prime }<0$ at each point of $\left(a,b\right)$ , then f decreases on $\left[a,b\right]$

It is assumed that f be continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ .

Calculation :

For decreasing function $y^{\prime }$ must be less than 0.

  $-\frac{1}{2\sqrt{x+2}}<0$

For each value of x inside the domain $\left[-2,\infty \right)$

  $\sqrt{x+2}$ is positive and it is multiplied by a negative sign therefore $y^{\prime }$ is always negative except for $x=-2$ .

Therefore,

The function decreases on $\left[-2,\infty \right)$