Question #fc572

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Answer 1 · 2017-05-02T00:45:57

There are two critical points, at #x = 0# and #x= -2#.

Since the derivative is already given, all we must do is solve the equation

#e^x(x^2 + 2x) = 0#

Step 1: solving #e^x = 0#

This has no solution, because if you take the natural logarithm of both sides, you get

#ln(e^x) = ln0#

And #ln(0)# has no real value.

Step 2: solving #x^2 + 2x = 0#

We have:

#x(x + 2) = 0#

#x = 0 or -2#

These are our two critical numbers.

Hopefully this helps!