What is the critical f(x)=xsqrt(36-x^2) ?

The critical f(x)=xsqrt(36-x^2) is x=-6,x=-3sqrt(2),x=3sqrt(2),x=6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical f(x)=xsqrt(36-x^2)

critical points

f (x) = x 36 - x^{2}

x = - 6, x = - 32, x = 32, x = 6

Solution steps

x 36 - x^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 32, x = 32, x = - 6, x = 6

Identify critical points not in

f (x)

domain

Domain of

x 36 - x^{2} : - 6 \leq x \leq 6

All critical points are in domain

x = - 6, x = - 32, x = 32, x = 6

x + 3 y = - 6

2 sin (2 x) + 3

f (x) = lo g_{4} (x - 2)

f (x) = - \frac{1}{3 - x} - 2

f (x) = \frac{2 x + 1}{4 - x}

What is the critical f(x)=xsqrt(36-x^2) ?
The critical f(x)=xsqrt(36-x^2) is x=-6,x=-3sqrt(2),x=3sqrt(2),x=6