# How do you find local maximum value of f using the first and second derivative tests: f(x)=x^3-2x+5 on the interval (-2,2)?

Nov 27, 2017

See below.

#### Explanation:

First differentiate ${x}^{3} - 2 x + 5$

$\frac{d}{\mathrm{dx}} \left({x}^{3} - 2 x + 5\right) = 3 {x}^{2} - 2$

equating this to zero and solving fot $x$ will give us the points where the tangent is horizontal .i.e. maximum and minimum values.

$3 {x}^{2} - 2 = 0 \implies x = \pm \sqrt{\frac{2}{3}}$

We now use the second derivative test. If:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$ minimum value.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0$ maximum value.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$ min/max or point of inflection.

Second derivative:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left(3 {x}^{2} - 2\right) = 6 x$

$6 \left(+ \sqrt{\frac{2}{3}}\right) > 0$ min value.

$6 \left(- \sqrt{\frac{2}{3}}\right) < 0$ max value.

These are both in the interval $\left(- 2 , 2\right)$

$x = - \sqrt{\frac{2}{3}}$ give maximum value so:

$f \left(- \sqrt{\frac{2}{3}}\right) = {\left(- \sqrt{\frac{2}{3}}\right)}^{3} - 2 \left(- \sqrt{\frac{2}{3}}\right) + 5$

$\to = \frac{4 \sqrt{6}}{9} + 5 \approx 6.089$

Graph of function and first derivative: