Question

Find all the critical numbers for the following function.
F(x) = (16 - x^2)^(3/5)

Answer 1

f' = (3/5)(16-x^2)^(-2/5)(-2x)
That is undefined when x = 4 or x = -4 and is zero when x = 0

Answer 2

f'(x) = (3/5)(16 - x^2)^(-2/5) (-2x)
= 0 for max/min of f(x)
-2x = 0
x = 0 , then y = 16^(3/5)

find f''(x) using the product rule
set that equal to zero to find any points of inflection

x-intercepts, let y = 0
(16-x^2)^(3/5) = 0
x^2 = 16
x = ± 4

y-intercept, let x = 0
y = 16^(3/5)