Question

find the critical points of the function (f(x)=x - 8ln(x)), for (x>0). use the first derivative test to determine whether (f) has a relative minimum, or maximum, or neither at its critical point(s). (separate multiple answers with commas. enter dne for \does not exist\.) all critical point(s) at (x =) (f) attains a relative minimum at (x =) (f) attains a relative maximum at (x =)

Explanation:

Step1: Find the first - derivative of the function

The function is $f(x)=x - 8\ln(x)$ with $x>0$. Using the sum - rule and the derivative formulas $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(\ln(x))=\frac{1}{x}$, we have $f^\prime(x)=\frac{d}{dx}(x)-8\frac{d}{dx}(\ln(x)) = 1-\frac{8}{x}=\frac{x - 8}{x}$.

Step2: Find the critical points

Critical points occur where $f^\prime(x)=0$ or $f^\prime(x)$ is undefined. Since $f^\prime(x)=\frac{x - 8}{x}$ and $x>0$, $f^\prime(x)$ is undefined when $x = 0$ but this is not in the domain of $f(x)$. Set $f^\prime(x)=0$:
$\frac{x - 8}{x}=0$. A fraction is zero when the numerator is zero and the denominator is non - zero. So $x−8 = 0$, which gives $x = 8$.

Step3: Use the First - Derivative Test

Choose test points in the intervals $(0,8)$ and $(8,\infty)$. Let's choose $x = 1$ for the interval $(0,8)$ and $x=9$ for the interval $(8,\infty)$.
For $x = 1$, $f^\prime(1)=\frac{1 - 8}{1}=-7<0$, so $f(x)$ is decreasing on the interval $(0,8)$.
For $x = 9$, $f^\prime(9)=\frac{9 - 8}{9}=\frac{1}{9}>0$, so $f(x)$ is increasing on the interval $(8,\infty)$.
Since $f(x)$ changes from decreasing to increasing at $x = 8$, $f(x)$ has a relative minimum at $x = 8$.

Answer:

All critical point(s) at $x = 8$
$f$ attains a relative minimum at $x = 8$
$f$ attains a relative maximum at $x=$ DNE