What is the extreme f(x)=sqrt(416-x)x ?

The extreme f(x)=sqrt(416-x)x is Maximum(832/3 ,(3328sqrt(26))/(3sqrt(3))),Minimum(416,0)

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extreme f(x)=sqrt(416-x)x

extreme points

f (x) = 416 - x x

Maximum (\frac{832}{3}, \frac{3328 26}{3 3}), Minimum (416, 0)

Solution steps

f^{'} (x) = \frac{- 3 x + 832}{2 416 - x}

Find intervals:

Increasing

: - \infty < x < \frac{832}{3},

Decreasing

: \frac{832}{3} < x < 416

Plug

x = \frac{832}{3}

into

416 - x x : \frac{3328 26}{3 3}

Maximum (\frac{832}{3}, \frac{3328 26}{3 3})

Plug

x = 416

into

416 - x x : 0

Minimum (416, 0)

Maximum (\frac{832}{3}, \frac{3328 26}{3 3}), Minimum (416, 0)

f (x) = x^{\frac{4}{5}} - 4

f (x) = (\frac{5 x ^{4}}{ln ( x )})

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

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

f (x) = 2 x^{3} - 6 x + 11

What is the extreme f(x)=sqrt(416-x)x ?
The extreme f(x)=sqrt(416-x)x is Maximum(832/3 ,(3328sqrt(26))/(3sqrt(3))),Minimum(416,0)