What is the extreme f(x)=-45x^2+400x ?

The extreme f(x)=-45x^2+400x is Maximum(40/9 , 8000/9)

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extreme f(x)=-45x^2+400x

extreme points

f (x) = - 45 x^{2} + 400 x

Maximum (\frac{40}{9}, \frac{8000}{9})

Solution steps

f^{'} (x) = - 90 x + 400

Find intervals:

Increasing

: - \infty < x < \frac{40}{9},

Decreasing

: \frac{40}{9} < x < \infty

Plug

x = \frac{40}{9}

into

- 45 x^{2} + 400 x : \frac{8000}{9}

Maximum (\frac{40}{9}, \frac{8000}{9})

f (x) = 3 x^{2} - 2 x + 5

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

f (x) = x^{3} + 8 x + 12

7 - 6 x - x^{3}

f (x) = x (e^{- \frac{x}{3}})

What is the extreme f(x)=-45x^2+400x ?
The extreme f(x)=-45x^2+400x is Maximum(40/9 , 8000/9)