How do you find the critical points of d'(x)=x^2+2x-4?

Jul 1, 2018

$x = - 1 \pm \sqrt{5}$

$\approx 1.236 , - 3.236$

Explanation:

We know that a function $f \left(x\right)$ has critical points wherever one of two conditions are met:

• $f ' \left(x\right) = 0$
• $f ' \left(x\right)$ is undefined.

So, let's look for such cases in our function here. We've been given the derivative $d ' \left(x\right)$, so we just need to set this equal to 0:

$d ' \left(x\right) = {x}^{2} + 2 x - 4 = 0$

Note that since $d ' \left(x\right)$ is a polynomial, and polynomials are defined everywhere, we don't need to worry about finding where $d ' \left(x\right)$ is undefined.

Since this doesn't factor, we'll need to use the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\implies x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 4\right)}}{2 \left(1\right)}$

$\implies x = \frac{- 2 \pm \sqrt{20}}{2}$

Some quick simplification/cleaning up gives:

$\implies x = - 1 \pm \sqrt{5}$

$\approx 1.236 , - 3.236$