# How do you find local maximum value of f using the first and second derivative tests: f(x) = x / (x^2 + 9)?

Apr 9, 2016

The first derivative yields a critical point at $x = 3$ i.e.
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 0 , \text{ for } {x}_{1 , 2} = \pm 3$
The second derivative $\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} ^ 2 {|}_{x = 3} = - 0.01852$

Since $\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} ^ 2 {|}_{x = 3} < 0$ we have local maximum

#### Explanation:

Given: $f \left(x\right) = \frac{x}{{x}^{2} + 9}$

Required: Local Maxima

Solution Strategy:

1) Take the 1st derivative to find the critical point, $c$, by solving
$f ' \left(x\right) = 0$. The solution is the critical point

2) Take the 2nd derivative and evaluate it at $c$:

• If (d^2f(x))/dx^2|_(x=c)>0; => "local minimum"
• else (d^2f(x))/dx^2|_(x=c)<0; => "local maximum"

1) Use quotient rule to differentiate f(x)

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{{x}^{2} + 9 - 2 {x}^{2}}{{x}^{2} + 9} ^ 2 = \frac{9 - {x}^{2}}{{x}^{2} + 9} ^ 2$

Now set $f ' \left(x\right) = 0$ and solve

$0 = \frac{9 - {x}^{2}}{{x}^{2} + 9} ^ 2$ so the critical points are ${x}_{1 , 2} = \pm 3$

2) $\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} ^ 2 = \frac{- 2 x {\left({x}^{2} + 9\right)}^{2} - \left(4 x \left({x}^{2} + 9\right)\right) \left(9 - {x}^{2}\right)}{{x}^{2} + 9} ^ 4$
$= 2 x \frac{{x}^{2} - 27}{{x}^{2} + 9} ^ 3 {|}_{x = 3} = - 0.01852$ Thus we have a maximum