What is the extreme f(x)=-300x^2+13500x-5400 ?

The extreme f(x)=-300x^2+13500x-5400 is Maximum(45/2 ,146475)

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extreme f(x)=-300x^2+13500x-5400

extreme points

f (x) = - 300 x^{2} + 13500 x - 5400

Maximum (\frac{45}{2}, 146475)

Solution steps

f^{'} (x) = - 600 x + 13500

Find intervals:

Increasing

: - \infty < x < \frac{45}{2},

Decreasing

: \frac{45}{2} < x < \infty

Plug

x = \frac{45}{2}

into

- 300 x^{2} + 13500 x - 5400 : 146475

Maximum (\frac{45}{2}, 146475)

f (x) = sin (x), 0 \leq x < \frac{π}{2}

f (x) = 3 - ∣ x - 2 ∣

f (x) = \frac{1}{4} x^{4} - \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 1

f (x, y) = \frac{3 x ^{2} - 1}{x ^{2} + y ^{2} + 1}

f (x) = 3 x + \frac{27}{x}

What is the extreme f(x)=-300x^2+13500x-5400 ?
The extreme f(x)=-300x^2+13500x-5400 is Maximum(45/2 ,146475)