What is the extreme f(x)=-0.001x^2+9x-1296 ?

The extreme f(x)=-0.001x^2+9x-1296 is Maximum(4500,18954)

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extreme f(x)=-0.001x^2+9x-1296

extreme points

f (x) = - 0.001 x^{2} + 9 x - 1296

Maximum (4500, 18954)

Solution steps

f^{'} (x) = - 0.002 x + 9

Find intervals:

Increasing

: - \infty < x < 4500,

Decreasing

: 4500 < x < \infty

Plug

x = 4500

into

- 0.001 x^{2} + 9 x - 1296 : 18954

Maximum (4500, 18954)

f (x) = \frac{( 500 + 500 \cdot x )}{1 + 0.04 \cdot x ^{2}}

\frac{( x ^{2} + 2 )}{( x ^{2} - 16 )}

y = 2 x^{2} + 5

f (x) = 5 x^{4} - 20 x^{3} + 1

f (x) = 0.001 x^{2} + 3.2 x - 10

What is the extreme f(x)=-0.001x^2+9x-1296 ?
The extreme f(x)=-0.001x^2+9x-1296 is Maximum(4500,18954)