What is the extreme f(x)=1125x-0.15x^3 ?

The extreme f(x)=1125x-0.15x^3 is Minimum(-50,-37500),Maximum(50,37500)

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extreme f(x)=1125x-0.15x^3

extreme points

f (x) = 1125 x - 0.15 x^{3}

Minimum (- 50, - 37500), Maximum (50, 37500)

Solution steps

f^{'} (x) = 1125 - 0.45 x^{2}

Find intervals:

Decreasing

: - \infty < x < - 50,

Increasing

: - 50 < x < 50,

Decreasing

: 50 < x < \infty

Plug

x = - 50

into

1125 x - 0.15 x^{3} : - 37500

Minimum (- 50, - 37500)

Plug

x = 50

into

1125 x - 0.15 x^{3} : 37500

Maximum (50, 37500)

Minimum (- 50, - 37500), Maximum (50, 37500)

y = \frac{x - 8}{( x - 3 ) ( x + 5 )}

f (x) = 1 - sin (2 x)

ln (x + 3)

\frac{x ^{4} + 2 x ^{2} + 1}{x ^{3} - 1}

ln (\frac{x - 9}{x - 7})

What is the extreme f(x)=1125x-0.15x^3 ?
The extreme f(x)=1125x-0.15x^3 is Minimum(-50,-37500),Maximum(50,37500)