What is the extreme f(x)=sqrt(5t^2+24t+136) ?

The extreme f(x)=sqrt(5t^2+24t+136) is Minimum(-12/5 ,(2sqrt(134))/(sqrt(5)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=sqrt(5t^2+24t+136)

extreme points

f (x) = 5 t^{2} + 24 t + 136

Minimum (- \frac{12}{5}, \frac{2 134}{5})

Solution steps

f^{'} (t) = \frac{5 t + 12}{5 t ^{2} + 24 t + 136}

Find intervals:

Decreasing

: - \infty < t < - \frac{12}{5},

Increasing

: - \frac{12}{5} < t < \infty

Plug

t = - \frac{12}{5}

into

5 t^{2} + 24 t + 136 : \frac{2 134}{5}

Minimum (- \frac{12}{5}, \frac{2 134}{5})

4 x^{3} - 3 x^{2} - 36 x + 17

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

f (x) = \frac{e ^{x}}{x ^{5}}

y = 7 - 6 x - x^{3}

f (x) = \frac{1}{x ^{3}}

What is the extreme f(x)=sqrt(5t^2+24t+136) ?
The extreme f(x)=sqrt(5t^2+24t+136) is Minimum(-12/5 ,(2sqrt(134))/(sqrt(5)))