# How do you find local maximum value of f using the first and second derivative tests: f(x) = 4x^3 + 3x^2 - 6x + 1?

Nov 9, 2016

$f ' ' \left(- 1\right) = 24 \cdot \left(- 1\right) + 6 = \left(- 18\right)$ - local maximum
$f ' ' \left(\frac{1}{2}\right) = 24 \cdot \left(\frac{1}{2}\right) + 6 = 18$ - local minimum

#### Explanation:

$f ' \left(x\right) = 12 \cdot {x}^{2} + 6 \cdot x - 6$
$f ' \left(x\right) = 0$
$12 {x}^{2} + 6 x - 6 = 0$
${x}_{1} = \left(- 1\right)$
${x}_{2} = \frac{1}{2}$

there are two values -local maximum ${x}_{1} , {x}_{2}$

$f ' ' \left(x\right) = 24 x + 6$

$f ' ' \left(- 1\right) = 24 \cdot \left(- 1\right) + 6 = \left(- 18\right)$ - local maximum, <0
$f ' ' \left(\frac{1}{2}\right) = 24 \cdot \left(\frac{1}{2}\right) + 6 = 18$ - local minimum, >0