# find the critical numbers of the function:

• October 26th 2009, 04:28 PM
hazecraze
find the critical numbers of the function:
$f(x)=x^\frac{4}{5}(x-4)^2$
find the critical numbers of the function:

$f'(x)=\frac{4}{5}x^\frac{-1}{5}(x-4)^2 + 2(x-4)(x^\frac{4}{5})
$

factor a term out:
$x^\frac{-1}{5}(\frac{4}{5}(x-4)^2 + 2x^\frac{5}{5}(x-4))$

I see the 0 and 4, but the solution guide says there is also a critical number at $\frac{8}{7}$.
• October 26th 2009, 04:57 PM
Debsta
(x-4) is a common factor so take it out the front too to get:
x^(-1/5)(x-4)[(4/5)(x-4)+2x]

Last bracket becomes (4/5)x-16/5+2x Set this =0 and solve to get the third solution.