# How do you find and classify local maxima, local minima, and all critical points of f(x) = x^3+x^2-4x-4?

Jul 27, 2018

There is a local maximum at $\left(- 1.535 ,\right) .879$ and a local minimum at $\left(0.869 , - 6.065\right)$. The point of inflection is $= \left(- \frac{1}{3} , - 2.593\right)$.

#### Explanation:

The function is

$f \left(x\right) = {x}^{3} + {x}^{2} - 4 x - 4$

As this is a polynomial function, the domain is $\mathbb{R}$

Calculate the first derivative

$f ' \left(x\right) = 3 {x}^{2} + 2 x - 4$

The critical points are when

$f ' \left(x\right) = 0$

That is

$3 {x}^{2} + 2 x - 4 = 0$

The solutions to this quadratic equation are

$x = \frac{- 2 \pm \sqrt{4 + 48}}{2 \cdot 3} = \frac{- 2 \pm \sqrt{52}}{6}$

Therefore,

${x}_{1} = \frac{- 2 - \sqrt{52}}{6}$

${x}_{2} = \frac{- 2 + \sqrt{52}}{6}$

To determine the nature of thr critical points, you can build either variation chart or calculate the second derivatives.

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$${x}_{1}$$\textcolor{w h i t e}{a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$↗$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a a}$↗

There is a local maximum at $\left(- 1.535 ,\right) .879$ and a local minimum at $\left(0.869 , - 6.065\right)$

Calculate the second derivative

$f ' ' \left(x\right) = 6 x + 2$

And the point of inflections when

$f ' ' \left(x\right) = 0$

$\implies$, $x = - \frac{1}{3}$

The point of inflection is $= \left(- \frac{1}{3} , - 2.593\right)$

graph{x^3+x^2-4x-4 [-10, 10, -5, 5]}

Point of maximum: $x = - 1.535$

Point of minimum: $x = 0.868$

Point of inflection: $x = - \frac{1}{3}$

#### Explanation:

Given function:

$f \left(x\right) = {x}^{3} + {x}^{2} - 4 x - 4$

$f ' \left(x\right) = 3 {x}^{2} + 2 x - 4$

$f ' ' \left(x\right) = 6 x + 2$

For maximum or minimum points, we must have

$f ' \left(x\right) = 0$

$\setminus \therefore 3 {x}^{2} + 2 x - 4 = 0$

$x = \setminus \frac{- 2 \setminus \pm \setminus \sqrt{{2}^{2} - 4 \left(3\right) \left(- 4\right)}}{2 \left(3\right)}$

$x = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{13}}{3}$

$= 0.868 , - 1.535$

$\setminus \implies f ' ' \left(0.868\right) = 6 \left(0.868\right) + 2 = 7.208 > 0$

hence, given function is minimum at $x = 0.868$

$\setminus \implies f ' ' \left(- 1.535\right) = 6 \left(- 1.535\right) + 2 = - 7.21 < 0$

hence, given function is maximum at $x = - 1.535$

Now, the point of inflection will occur at $f ' ' \left(x\right) = 0$

$\setminus \therefore 6 x + 2 = 0$

$x = - \frac{1}{3}$