What is the extreme f(x)=120x-0.4x^4+800 ?

The extreme f(x)=120x-0.4x^4+800 is Maximum(\sqrt[3]{75},1179.54469…)

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extreme f(x)=120x-0.4x^4+800

extreme points

f (x) = 120 x - 0.4 x^{4} + 800

Solution steps

f^{'} (x) = - 1.6 x^{3} + 120

Find intervals:

Increasing

Decreasing

Plug

into

120 x - 0.4 x^{4} + 800 : 1179.54469 \dots

(- 2, 3), (4, 5)

f (x) = 5 x - 30

f (x) = t^{4} - 16 t^{3} + 64 t^{2}

(2, - 7), (9, - 2)

f (x) = lo g_{5} (x^{3})

What is the extreme f(x)=120x-0.4x^4+800 ?
The extreme f(x)=120x-0.4x^4+800 is Maximum(\sqrt[3]{75},1179.54469…)