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 :)