Question

find the $x$-coordinate of all relative maximums and minimums of $f(x)=e^{2x}(x - 4)^9$. use exact values. maximum at $x=$ minimum at $x=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = e^{2x}$ and $v=(x - 4)^9$. Then $u^\prime=2e^{2x}$ and $v^\prime = 9(x - 4)^8$. So $f^\prime(x)=2e^{2x}(x - 4)^9+9e^{2x}(x - 4)^8$.

Step2: Factor out common terms

Factor out $e^{2x}(x - 4)^8$ from $f^\prime(x)$. We get $f^\prime(x)=e^{2x}(x - 4)^8(2(x - 4)+9)$.

Step3: Simplify the expression

Expand and simplify $2(x - 4)+9$ to get $2x-8 + 9=2x + 1$. So $f^\prime(x)=e^{2x}(x - 4)^8(2x + 1)$.

Step4: Find critical points

Set $f^\prime(x)=0$. Since $e^{2x}>0$ for all real $x$, we set $(x - 4)^8(2x + 1)=0$. Then $x-4 = 0$ or $2x+1 = 0$. So the critical points are $x = 4$ and $x=-\frac{1}{2}$.

Step5: Use the first - derivative test

Choose test points in the intervals $(-\infty,-\frac{1}{2})$, $(-\frac{1}{2},4)$ and $(4,\infty)$.
For the interval $(-\infty,-\frac{1}{2})$, let $x=-1$. Then $f^\prime(-1)=e^{-2}(-1 - 4)^8(-2 + 1)<0$.
For the interval $(-\frac{1}{2},4)$, let $x = 0$. Then $f^\prime(0)=e^{0}(0 - 4)^8(0 + 1)>0$.
For the interval $(4,\infty)$, let $x = 5$. Then $f^\prime(5)=e^{10}(5 - 4)^8(10 + 1)>0$.

Since the function changes from decreasing to increasing at $x=-\frac{1}{2}$, there is a relative minimum at $x =-\frac{1}{2}$. Since the function does not change its increasing - decreasing behavior at $x = 4$ (the sign of the derivative is non - negative on both sides of $x = 4$), there is no relative maximum or minimum at $x = 4$.

Answer:

maximum at $x=\text{None}$
minimum at $x=-\frac{1}{2}$