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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