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Answer 1 · 2017-05-02T00:45:57

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

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

$e^{x} (x^{2} + 2 x) = 0$ $e^x(x^2 + 2x) = 0$

Step 1: solving $e^{x} = 0$ $e^x = 0$

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

$ln (e^{x}) = ln 0$ $ln(e^x) = ln0$

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

Step 2: solving $x^{2} + 2 x = 0$ $x^2 + 2x = 0$

We have:

$x (x + 2) = 0$ $x(x + 2) = 0$

$x = 0 or - 2$ $x = 0 or -2$

These are our two critical numbers.

Hopefully this helps!