a. find the open interval(s) on which the function is increasing and decreasing.\nb. identify the functions…

a. find the open interval(s) on which the function is increasing and decreasing.\nb. identify the functions local and absolute extreme values, if any, saying where they occur.\nf(x)=10x ln x\nd. there are no local maxima.\nfind each local minimum, if there are any.\n(type exact answers.)\na. the function has a local minimum value at two values of x. in increasing order of x - value, the minimum values are f() = and f() =\nb. the function has a local minimum value at three values of x. in increasing order of x - value, the minimum values are f() =, f() =, and f() =\nc. the function has a local minimum value at one value of x. the minimum value is f() =\nd. there are no local minima.

a. find the open interval(s) on which the function is increasing and decreasing.\nb. identify the functions local and absolute extreme values, if any, saying where they occur.\nf(x)=10x ln x\nd. there are no local maxima.\nfind each local minimum, if there are any.\n(type exact answers.)\na. the function has a local minimum value at two values of x. in increasing order of x - value, the minimum values are f() = and f() =\nb. the function has a local minimum value at three values of x. in increasing order of x - value, the minimum values are f() =, f() =, and f() =\nc. the function has a local minimum value at one value of x. the minimum value is f() =\nd. there are no local minima.

Answer

Explanation:

Step1: Find the derivative of the function

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 10x) and (v=\ln x). (u^\prime=10), (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), so (x = e^{-1}=\frac{1}{e}). The domain of (y = f(x)) is ((0,+\infty)).

Step3: Determine the intervals of increase and decrease

Take a test - point in the interval ((0,\frac{1}{e})), say (x=\frac{1}{e^{2}}). (f^\prime(\frac{1}{e^{2}})=10\ln(\frac{1}{e^{2}})+10=10(-2)+10=-10<0). Take a test - point in the interval ((\frac{1}{e},+\infty)), say (x = 1). (f^\prime(1)=10\ln(1)+10=10>0).

So the function (y = f(x)) is decreasing on the interval ((0,\frac{1}{e})) and increasing on the interval ((\frac{1}{e},+\infty)).

Step4: Find the local extreme values

Since the function changes from decreasing to increasing at (x=\frac{1}{e}), by the First - Derivative Test, there is a local minimum at (x = \frac{1}{e}). (f(\frac{1}{e})=10\times\frac{1}{e}\ln(\frac{1}{e})=-\frac{10}{e}).

Answer:

a. The function (f(x)=10x\ln x) is decreasing on the interval ((0,\frac{1}{e})) and increasing on the interval ((\frac{1}{e},+\infty)). b. The function has a local minimum value. The local minimum value is (f(\frac{1}{e})=-\frac{10}{e}). There are no local maxima. So for the local - minimum part, the answer is C. The function has a local minimum value at one value of (x). The minimum value is (f(\frac{1}{e})=-\frac{10}{e}).