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
(type your answer in interval notation. type exact answers. use a comma to separate answers as needed.)
○ b. the function is never increasing.
on what open interval(s), if any, is the function decreasing? select the correct choice below and fill in any answer boxes within your choice.
● a. the function f is decreasing on the open interval(s) (-∞, -1/e) (type your answer in interval notation. type exact answers. use a comma to separate answers as needed.)
○ b. the function is never decreasing

Explanation:

Step1: Find the derivative of the function

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

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(10\ln x+10 = 0\).
\(\ln x=- 1\), and \(x = e^{-1}=\frac{1}{e}\).
The domain of \(y = f(x)\) is \((0,\infty)\) (since \(\ln x\) is defined for \(x>0\)).

Step3: Determine the intervals of increase and decrease

We use test - points.
For the interval \((0,\frac{1}{e})\), let \(x=\frac{1}{e^{2}}\). Then \(f^\prime(\frac{1}{e^{2}})=10\ln(\frac{1}{e^{2}})+10=10(-2)+10=-10<0\).
For the interval \((\frac{1}{e},\infty)\), let \(x = 1\). Then \(f^\prime(1)=10\ln(1)+10=10>0\).

Answer:

The function \(f(x)=10x\ln x\) is decreasing on the open interval \((0,\frac{1}{e})\) and increasing on the open interval \((\frac{1}{e},\infty)\).