Respuesta :
Answer:
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!!!
