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

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

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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})

f (x) = \frac{2 x + 9}{x - 3}

f (x) = x + 2 + 3

f (x) = 5 x - 7

y = \frac{e ^{x}}{1 + 2 e ^{x}}

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

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