Question

attempt 1: 10 attempts remaining. find the critical points of the function f(x)=x - 3ln(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” if either a critical point or relative minimum/maximum 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 derivative of the function

The derivative of $y = x$ is $y^\prime=1$ and the derivative of $y = 3\ln(x)$ is $y^\prime=\frac{3}{x}$ using the power - rule and the derivative formula for $\ln(x)$. So, $f^\prime(x)=1-\frac{3}{x}=\frac{x - 3}{x}$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. So, $\frac{x - 3}{x}=0$. This implies $x-3 = 0$ (since $x>0$), and we get $x = 3$. The derivative is undefined when $x = 0$, but our domain is $x>0$, so the only critical point is $x = 3$.

Step3: Apply the First - Derivative Test

Choose test points in the intervals $(0,3)$ and $(3,\infty)$. Let's choose $x = 1$ for the interval $(0,3)$ and $x=4$ for the interval $(3,\infty)$.
When $x = 1$, $f^\prime(1)=\frac{1 - 3}{1}=-2<0$, so the function is decreasing on the interval $(0,3)$.
When $x = 4$, $f^\prime(4)=\frac{4 - 3}{4}=\frac{1}{4}>0$, so the function is increasing on the interval $(3,\infty)$.

Answer:

All critical point(s) at $x = 3$
$f$ attains a relative minimum at $x = 3$
$f$ attains a relative maximum at $x = DNE$