Answer the following questions about the function whose derivative is f prime(x)equalsx Superscript negative one third Baseline (x minus 5 ). a. What are the critical points of f? b. On what open intervals is f increasing or decreasing? c. At what points, if any, does f assume local maximum and minimum values? a. Find the critical points, if any. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a) The critical points for the function include x = 0 and x = 5.

b) The function is increasing when x < 0 and when x > 5.

The function is decreasing when 0 < x < 5.

c) The function has a local minimum at x=5, no known maximum point could be obtained though (no local maximum exists).

Step-by-step explanation:

f'(x) = x⁻⁰•³³³³ (x - 5)

At critical points,

f'(x) = 0

x⁻⁰•³³³³(x - 5) = 0

x⁻⁰•³³³³ = 0 or (x-5) = 0

x = 0 or x = 5

b) A function is said to be increasing when

f'(x) > 0.

and it is said to be decreasing when

f'(x) < 0.

To investigate the region's of increasing or decreasing function.

f'(x) = x⁻⁰•³³³³ (x - 5)

with solutions of x=0 and x=5 at critical points. So, we check the behaviour of f'(x) around the critical points.

x < 0, 0 < x < 5, x > 5

x < 0 | 0 < x < 5 | x > 5 | function

-ve positive ||| +ve | (x⁻⁰•³³³³)

-ve negative || +ve | (x - 5)

+ve | negative | +ve | x⁻⁰•³³³³ (x - 5)

f'(x) = x⁻⁰•³³³(x - 5) > 0 when x < 0 and x > 5.

f'(x) = x⁻⁰•³³³³ (x - 5) < 0 when 0 < x < 5.

c) At maximum point, f"(x) < 0

And at minimum point, f"(x) > 0

f'(x) = x⁻⁰•³³³³ (x - 5) = x⁰•⁶⁶⁷ - 5x⁻⁰•³³³³

f'(x) = x⁰•⁶⁶⁷ - 5x⁻⁰•³³³³

f"(x) = (2/3) x⁻⁰•³³³³ - (-5/3)x⁻¹•³³³³

f"(x) = (2/3) x⁻⁰•³³³³ + (5/3)x⁻¹•³³³³

So, inserting the critical points, 0 and 5

At x = 0,

f"(x) = (2/3) (0)⁻⁰•³³³³ + (5/3)(0)⁻¹•³³³³ = 0.

This means that, the point x = 0 isn't a maximum or minimum point.

At x = 5

f"(x) = (2/3) (5)⁻⁰•³³³³ + (5/3)(5)⁻¹•³³³³

= 0.390 + 0.195 = 0.585 > 0

This corresponds to a minimum point.

Hope this Helps!!!

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50