# How do you find local maximum value of f using the first and second derivative tests: f(x) = x + sqrt(9 − x) ?

Aug 28, 2016

$f \left(\frac{35}{4}\right)$ is a local maximum of $\frac{37}{4}$

#### Explanation:

$f \left(x\right) = x + \sqrt{9 - x}$

$f ' \left(x\right) = 1 + \frac{1}{2} {\left(9 - x\right)}^{- \frac{1}{2}} \cdot \left(- 1\right) = 1 - \frac{1}{2 \sqrt{9 - x}}$

For local maxima or minima $f ' \left(x\right) = 0$

$f ' \left(x\right) = 0 \to 1 - \frac{1}{2 \sqrt{9 - x}} = 0$

$\frac{1}{\sqrt{9 - x}} = 2$

$\sqrt{9 - x} = \frac{1}{2}$

$9 - x = \frac{1}{4}$

$x = 9 - \frac{1}{4} = \frac{35}{4}$
$\to f \left(x\right) = \frac{35}{4} + \sqrt{9 - \frac{35}{4}}$

$= \frac{35}{4} + \sqrt{\frac{1}{4}}$

$= \frac{35}{4} + \frac{1}{2} = \frac{37}{4}$

Hence $f \left(x\right)$ has a turning point at $\left(\frac{35}{4} , \frac{37}{4}\right)$

Now consider $f ' ' \left(x\right) = 0 - \frac{d}{\mathrm{dx}} \frac{1}{2} {\left(9 - x\right)}^{- \frac{1}{2}}$

$= - \left(- \frac{1}{4} {\left(9 - x\right)}^{- \frac{3}{2}} \cdot \left(- 1\right)\right)$

$= - \frac{1}{4 {\left(9 - x\right)}^{\frac{3}{2}}}$

$f ' ' \left(\frac{35}{4}\right) = - \frac{1}{4 {\left(9 - \frac{35}{4}\right)}^{\frac{3}{2}}} = - \frac{1}{4 \cdot {\left(\frac{1}{4}\right)}^{\frac{3}{2}}}$

Hence $f ' ' \left(\frac{35}{4}\right) < 0 \to f \left(\frac{35}{4}\right)$ is a local maximum = $\frac{37}{4}$

This may be seen by the graph of $f \left(x\right)$ in the region of $f \left(\frac{35}{4}\right)$:

graph{x+sqrt(9-x) [7.9024, 9.588, 8.735, 9.5774]}