What is the extreme f(x)=-16t^2+60t+2 ?

The extreme f(x)=-16t^2+60t+2 is Maximum(15/8 , 233/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=-16t^2+60t+2

extreme points

f (x) = - 16 t^{2} + 60 t + 2

Maximum (\frac{15}{8}, \frac{233}{4})

Solution steps

f^{'} (t) = - 32 t + 60

Find intervals:

Increasing

: - \infty < t < \frac{15}{8},

Decreasing

: \frac{15}{8} < t < \infty

Plug

t = \frac{15}{8}

into

- 16 t^{2} + 60 t + 2 : \frac{233}{4}

Maximum (\frac{15}{8}, \frac{233}{4})

f (x) = \frac{- 4 x}{5 + 3 x}

f (x) = \frac{( x + 17 )}{( x - 14 )}

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

y = x^{2} - 7

f (x) = - x^{4} - 7 x^{6} + 2 x^{2}

What is the extreme f(x)=-16t^2+60t+2 ?
The extreme f(x)=-16t^2+60t+2 is Maximum(15/8 , 233/4)