What is the extreme f(x)=-16t^2+65t+6 ?

The extreme f(x)=-16t^2+65t+6 is Maximum(65/32 , 4609/64)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=-16t^2+65t+6

extreme points

f (x) = - 16 t^{2} + 65 t + 6

Maximum (\frac{65}{32}, \frac{4609}{64})

Solution steps

f^{'} (t) = - 32 t + 65

Find intervals:

Increasing

: - \infty < t < \frac{65}{32},

Decreasing

: \frac{65}{32} < t < \infty

Plug

t = \frac{65}{32}

into

- 16 t^{2} + 65 t + 6 : \frac{4609}{64}

Maximum (\frac{65}{32}, \frac{4609}{64})

What is the extreme f(x)=-16t^2+65t+6 ?
The extreme f(x)=-16t^2+65t+6 is Maximum(65/32 , 4609/64)