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

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Answer 1 · 2018-07-01T19:58:51

$x = -1 +- sqrt(5)$

$~~ 1.236, -3.236$

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

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

$d'(x) = x^2 + 2x - 4 = 0$

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

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

$x = [-b +- sqrt(b^2 - 4ac)]/(2a)$

$=> x = [-2 +- sqrt(2^2 - 4(1)(-4))]/(2(1))$

$=> x = [-2+-sqrt(20)]/2$

Some quick simplification/cleaning up gives:

$=> x = -1 +- sqrt(5)$

$~~ 1.236, -3.236$

And there's your answers.

Hope that helped :)

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