What is the extreme-1/80 t^2+9/40 t-17/80 ?

The extreme-1/80 t^2+9/40 t-17/80 is Maximum(9, 4/5)

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extreme-1/80 t^2+9/40 t-17/80

extreme points

- \frac{1}{80} t^{2} + \frac{9}{40} t - \frac{17}{80}

Maximum (9, \frac{4}{5})

Solution steps

f^{'} (t) = \frac{- t + 9}{40}

Find intervals:

Increasing

: - \infty < t < 9,

Decreasing

: 9 < t < \infty

Plug

t = 9

into

- \frac{1}{80} t^{2} + \frac{9}{40} t - \frac{17}{80} : \frac{4}{5}

Maximum (9, \frac{4}{5})

3 x^{2} - 4 x - 1

f (x) = f (x) = 3 x \frac{5}{3} - 15 x \frac{2}{3}

\frac{x + 3}{x ^{3} + 2 x ^{2} - 15 x - 36}

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

f (x, y) = (x - 3) (y - 3) (x + y - 3)

What is the extreme-1/80 t^2+9/40 t-17/80 ?
The extreme-1/80 t^2+9/40 t-17/80 is Maximum(9, 4/5)