What is the extreme f(x)=-0.002x^2+3.2x-60 ?

The extreme f(x)=-0.002x^2+3.2x-60 is Maximum(800,1220)

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extreme f(x)=-0.002x^2+3.2x-60

extreme points

f (x) = - 0.002 x^{2} + 3.2 x - 60

Maximum (800, 1220)

Solution steps

f^{'} (x) = - 0.004 x + 3.2

Find intervals:

Increasing

: - \infty < x < 800,

Decreasing

: 800 < x < \infty

Plug

x = 800

into

- 0.002 x^{2} + 3.2 x - 60 : 1220

Maximum (800, 1220)

f (x) = x^{3} + 6 x^{2} + 12 x + 3

f (x) = 7 x^{4} - 28 x^{3} + 5

f (x) = x^{4} - 2 4^{2} + 180

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

f (x) = x^{2} - 8 x, 0 \leq x \leq 9

What is the extreme f(x)=-0.002x^2+3.2x-60 ?
The extreme f(x)=-0.002x^2+3.2x-60 is Maximum(800,1220)