What is the critical h(t)=50+40t^3-5t^4-4t^5 ?

The critical h(t)=50+40t^3-5t^4-4t^5 is t=-3,t=0,t=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical h(t)=50+40t^3-5t^4-4t^5

critical points

h (t) = 50 + 40 t^{3} - 5 t^{4} - 4 t^{5}

t = - 3, t = 0, t = 2

Solution steps

50 + 40 t^{3} - 5 t^{4} - 4 t^{5}

Find where

f^{'} (t)

is equal to zero or undefined

t = - 3, t = 0, t = 2

Identify critical points not in

f (t)

domain

Domain of

50 + 40 t^{3} - 5 t^{4} - 4 t^{5} : - \infty < t < \infty

All critical points are in domain

t = - 3, t = 0, t = 2

R (t) = 1 + 80 t^{3} + 5 t^{4} - 2 t^{5}

f (x) = \frac{x ^{3} - 4 x ^{2} + 7 x}{x}

f (x) = x^{3} - 6 x + 1

f (x) = x^{2} + 6 x + 10

f (x) = I n (x - 4)

What is the critical h(t)=50+40t^3-5t^4-4t^5 ?
The critical h(t)=50+40t^3-5t^4-4t^5 is t=-3,t=0,t=2