What is the critical x^8(x-4)^7 ?

The critical x^8(x-4)^7 is x=0,x= 32/15 ,x=4

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critical x^8(x-4)^7

critical points

x^{8} (x - 4)^{7}

x = 0, x = \frac{32}{15}, x = 4

Solution steps

x^{8} (x - 4)^{7}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = \frac{32}{15}, x = 4

Identify critical points not in

f (x)

domain

Domain of

x^{8} (x - 4)^{7} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{32}{15}, x = 4

y = 3^{x}

2 (x - 4)^{2} + 3

f (x) = 4

(- 9, - 6), (- 2, - 2)

4 x^{3} + 48 x^{2} + 6 x + 3