What is the extreme f(x)=615t-15/5 t^2 ?

The extreme f(x)=615t-15/5 t^2 is Maximum(205/2 , 126075/4)

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extreme f(x)=615t-15/5 t^2

extreme points

f (x) = 615 t - \frac{15}{5} t^{2}

Maximum (\frac{205}{2}, \frac{126075}{4})

Solution steps

f^{'} (t) = 615 - 6 t

Find intervals:

Increasing

: - \infty < t < \frac{205}{2},

Decreasing

: \frac{205}{2} < t < \infty

Plug

t = \frac{205}{2}

into

615 t - \frac{15}{5} t^{2} : \frac{126075}{4}

Maximum (\frac{205}{2}, \frac{126075}{4})

f (x) = \frac{2 x ^{2} + 1}{x ^{2} - 4}

f (t) = t \cdot e^{- ∣ x ∣}

f (x) = \frac{x ^{4}}{4} + x^{3} - 14 x^{2} - 60 x

f (x) = 4 x + z

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 70

What is the extreme f(x)=615t-15/5 t^2 ?
The extreme f(x)=615t-15/5 t^2 is Maximum(205/2 , 126075/4)