Chapter 4.3, Problem 17E

To determine

To find : The local maximum and minimum values using both firs and second derivative test.

Explanation of Solution

Given: The function is $f\left(x\right)=x+\sqrt{1-x}$

Find the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=1-\frac{1}{2\sqrt{1-x}}\\ =\frac{2\sqrt{1-x}-1}{2\sqrt{1-x}}\\ 0=2\sqrt{1-x}-1\\ x=\frac{1}{4},\frac{3}{4}\end{array}$

And $f^{\prime }\left(x\right)$ does not exist when $x=1$ .

Thus, the critical points are $\frac{3}{4},1$

The function $f^{\prime }$ changes from positive to negative at $x=\frac{3}{4}$ .

So, local maximum value is $\left(\frac{3}{4},f\left(\frac{3}{4}\right)\right)=\left(\frac{3}{4},\frac{5}{4}\right)$

By second derivative method.

  $\begin{array}{c}{f}^{″}\left(x\right)=\frac{-1}{2}\left(\frac{-1}{2}\right)\left(-1\right)\cdot \frac{1}{\left(1-x\right)\sqrt{1-x}}\\ =\frac{1}{4\left(1-x\right)\sqrt{1-x}}\\ f^{″}\left(\frac{3}{4}\right)=\frac{1}{4\left(1-\frac{3}{4}\right)\sqrt{1-\frac{3}{4}}}\\ =-2\end{array}$

Since,

  $f^{″}\left(\frac{3}{4}\right)<0$ .

So, $f\left(x\right)$ has a local maximum at $x=\frac{3}{4}$ .

So, local maximum value is $\left(\frac{3}{4},f\left(\frac{3}{4}\right)\right)=\left(\frac{3}{4},\frac{5}{4}\right)$

First derivative method is slightly easier to apply in this case.