What is the extreme f(x)=-5t^2+61t+18 ?

The extreme f(x)=-5t^2+61t+18 is Maximum(61/10 , 4081/20)

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extreme f(x)=-5t^2+61t+18

extreme

Maximum (\frac{61}{10}, \frac{4081}{20})

Solution steps

f^{'} (t) = - 10 t + 61

Find intervals:

Increasing

: - \infty < t < \frac{61}{10},

Decreasing

: \frac{61}{10} < t < \infty

Plug

t = \frac{61}{10}

into

- 5 t^{2} + 61 t + 18 : \frac{4081}{20}

Maximum (\frac{61}{10}, \frac{4081}{20})

What is the extreme f(x)=-5t^2+61t+18 ?
The extreme f(x)=-5t^2+61t+18 is Maximum(61/10 , 4081/20)