Question

Which table identifies the x-values, if any, of the relative extrema of f(x) = (x^3 – 5x^2 + 11x – 11)e^x

Answer 1

Answer:

Relative minimum: x=0

Relative maximum: None

Step-by-step explanation:

Relative extrema are going to be located where the first derivative changes sign. Therefore, we must first find the derivative of the function:

$f'(x)=\frac{d}{dx}[(x^3-5x^2+11x-11)e^x]\\ \\f'(x)=(x^3-5x^2+11x-11)*\frac{d}{dx}(e^x)+[\frac{d}{dx}(x^3-5x^2+11x-11)]*e^x\\ \\f'(x)=(x^3-5x^2+11x-11)e^x+(3x^2-10x+11)e^x\\\\f'(x)=(x^3-2x^2+x)e^x\\\\f'(x)=xe^x(x^2-2x+1)\\\\f'(x)=xe^x(x-1)^2$

Next, we set $f'(x)=0$ where we determine our critical points:

$0=xe^x(x-1)^2$

$x=0,x=1$

We then test points around the critical points to find where the derivative changes sign. I will use the points $x=1$, $x=\frac{1}{2}$, and $x=\frac{3}{2}$:

$f'(-1)=(-1)e^{-1}((-1)-1)^2=-\frac{1}{e}(-2)^2=-\frac{4}{e}<0\\\\f'(\frac{1}{2})=\frac{1}{2}e^\frac{1}{2}(\frac{1}{2}-2)^2=\frac{\sqrt{e}}{2}(-\frac{1}{2})^2=\frac{\sqrt{e}}{8}>0\\\\f'(\frac{3}{2})=\frac{3}{2}e^\frac{3}{2}(\frac{3}{2}-2)^2=\frac{3\sqrt{e^3} }{2}(\frac{1}{2})^2=\frac{3\sqrt{e^3}}{8}>0$

As you can see, the derivative changes sign from negative to positive at $x=0$, but the sign stays positive at $x=1$. Therefore, the only critical point that is an extreme point is $x=0$, which is a relative minimum. This means that there is no relative maximum.

In conclusion, the 4th option is correct. Review the graph for a visual of how the derivative sign changes.