What is the critical f(x)=0.09x^2+16x+350 ?

The critical f(x)=0.09x^2+16x+350 is x=-800/9

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critical f(x)=0.09x^2+16x+350

critical points

f (x) = 0.09 x^{2} + 16 x + 350

x = - \frac{800}{9}

Solution steps

0.09 x^{2} + 16 x + 350

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{800}{9}

Identify critical points not in

f (x)

domain

Domain of

0.09 x^{2} + 16 x + 350 : - \infty < x < \infty

All critical points are in domain

x = - \frac{800}{9}

12 x^{3}

x - 1

f (x) = \frac{x ^{3} - 27}{x ^{2} - 7 x + 12}

5 x - y = 9

f (x) = x^{\frac{2}{3}} (x - 5)

What is the critical f(x)=0.09x^2+16x+350 ?
The critical f(x)=0.09x^2+16x+350 is x=-800/9