What is the extreme f(x)=-0.12x^2+168x+200 ?

The extreme f(x)=-0.12x^2+168x+200 is Maximum(700,59000)

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extreme f(x)=-0.12x^2+168x+200

extreme points

f (x) = - 0.12 x^{2} + 168 x + 200

Maximum (700, 59000)

Solution steps

f^{'} (x) = - 0.24 x + 168

Find intervals:

Increasing

: - \infty < x < 700,

Decreasing

: 700 < x < \infty

Plug

x = 700

into

- 0.12 x^{2} + 168 x + 200 : 59000

Maximum (700, 59000)

f (x) = \frac{ln ( x )}{x ^{4}}

\frac{2 x ^{2}}{x ^{2} - 9}

x^{3} ln (x)

f (x) = x^{3} - 6 x^{2} - 15 x + 1

\frac{4 x ^{2}}{x ^{2} - 4}

What is the extreme f(x)=-0.12x^2+168x+200 ?
The extreme f(x)=-0.12x^2+168x+200 is Maximum(700,59000)