Question

a. find the open interval(s) on which the function is increasing and decreasing
b. identify the functions local and absolute extreme values, if any, saying where they occur

f(x)=10x ln x

d. there are no local minima.

if the function has extreme values, which of the extreme values, if any, are absolute? select the correct
choice below and fill in any answer boxes within your choice.

(type exact answers. use a comma to separate answers as needed.)

a. there is no absolute maximum, but there is an absolute minimum of at x=

b. there is an absolute maximum of at x=, but no absolute minimum

c. there is an absolute maximum of at x= and an absolute minimum of at x=

d. there are local extreme values but there are no absolute extreme values

e. there are no local or absolute extreme values

Explanation:

Step1: Find the derivative

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 10x\) and \(v=\ln x\). Then \(u^\prime=10\) and \(v^\prime=\frac{1}{x}\). So \(f^\prime(x)=10\ln x + 10x\times\frac{1}{x}=10(\ln x + 1)\).

Step2: Find critical points

Set \(f^\prime(x)=0\), i.e., \(10(\ln x + 1)=0\). Then \(\ln x=- 1\), and \(x = e^{-1}=\frac{1}{e}\). The domain of \(f(x)\) is \((0,+\infty)\).

Step3: Determine increasing and decreasing intervals

• For \(x\in(0,\frac{1}{e})\), let \(x=\frac{1}{e^{2}}\). Then \(f^\prime(\frac{1}{e^{2}})=10(\ln\frac{1}{e^{2}}+1)=10(-2 + 1)=-10<0\). So \(f(x)\) is decreasing on \((0,\frac{1}{e})\).
• For \(x\in(\frac{1}{e},+\infty)\), let \(x = 1\). Then \(f^\prime(1)=10(\ln1 + 1)=10(0 + 1)=10>0\). So \(f(x)\) is increasing on \((\frac{1}{e},+\infty)\).

Step4: Find local and absolute extreme values

Since \(f(x)\) changes from decreasing to increasing at \(x = \frac{1}{e}\), \(f(\frac{1}{e})=10\times\frac{1}{e}\ln\frac{1}{e}=-\frac{10}{e}\) is a local minimum.
As \(x
ightarrow0^{+}\), \(f(x)=10x\ln x
ightarrow0\) (using L - H rule: \(\lim_{x
ightarrow0^{+}}10x\ln x=\lim_{x
ightarrow0^{+}}\frac{10\ln x}{\frac{1}{x}}\), and by L - H \(\lim_{x
ightarrow0^{+}}\frac{\frac{10}{x}}{-\frac{1}{x^{2}}}=\lim_{x
ightarrow0^{+}}(- 10x)=0\)). As \(x
ightarrow+\infty\), \(f(x)=10x\ln x
ightarrow+\infty\). So the local minimum \(f(\frac{1}{e})=-\frac{10}{e}\) is also an absolute minimum.

Answer:

A. There is no absolute maximum, but there is an absolute minimum of \(-\frac{10}{e}\) at \(x=\frac{1}{e}\)