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 points

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

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})

f (x, y) = x^{2} + y^{2} - 30

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x + 1

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

s (t) = - t^{2} + 5 t + 1

f (x) = x^{3} + 3 x^{4}

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