What is the extreme f(x)=(-125)/4 x^4+x ?

The extreme f(x)=(-125)/4 x^4+x is Maximum(1/5 , 3/20)

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extreme f(x)=(-125)/4 x^4+x

extreme points

f (x) = \frac{- 125}{4} x^{4} + x

Maximum (\frac{1}{5}, \frac{3}{20})

Solution steps

f^{'} (x) = - 125 x^{3} + 1

Find intervals:

Increasing

: - \infty < x < \frac{1}{5},

Decreasing

: \frac{1}{5} < x < \infty

Plug

x = \frac{1}{5}

into

\frac{- 125}{4} x^{4} + x : \frac{3}{20}

Maximum (\frac{1}{5}, \frac{3}{20})

f (x) = ln (x^{2} - 6 x - 7)

f (x) = sin (2 π x)

f (x) = x^{2} + y + x y

y = 2 + 3 x - x^{3}

f (x) = x^{2} + 9 x + 9

What is the extreme f(x)=(-125)/4 x^4+x ?
The extreme f(x)=(-125)/4 x^4+x is Maximum(1/5 , 3/20)