Posted by **Jordan** on Wednesday, January 18, 2012 at 3:49pm.

Find all the critical numbers for the following function.

F(x) = (16 - x^2)^(3/5)

- Calculus (Help please) -
**Damon**, Wednesday, January 18, 2012 at 5:08pm
f' = (3/5)(16-x^2)^(-2/5)(-2x)

That is undefined when x = 4 or x = -4 and is zero when x = 0

- Calculus (Help please) -
**Reiny**, Wednesday, January 18, 2012 at 5:13pm
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)

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