What is the extreme f(x)=-32x+36x^{1/2}+24 ?

The extreme f(x)=-32x+36x^{1/2}+24 is Maximum(0,24),Minimum(81/256 , 273/8)

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extreme f(x)=-32x+36x^{1/2}+24

extreme points

f (x) = - 32 x + 36 x^{\frac{1}{2}} + 24

Maximum (0, 24), Minimum (\frac{81}{256}, \frac{273}{8})

Solution steps

f^{'} (x) = \frac{18}{x ^{\frac{1}{2}}} - 32

Find intervals:

Decreasing

: 0 \leq x < \frac{256}{81},

Increasing

: \frac{256}{81} < x < \infty

Plug

x = 0

into

- 32 x + 36 x^{\frac{1}{2}} + 24 : 24

Maximum (0, 24)

Plug

x = \frac{81}{256}

into

- 32 x + 36 x^{\frac{1}{2}} + 24 : \frac{273}{8}

Minimum (\frac{81}{256}, \frac{273}{8})

Maximum (0, 24), Minimum (\frac{81}{256}, \frac{273}{8})

f (x) = \frac{x - 3}{2 x ^{2} - 3 x - 20}

8 x - 7 y - 14 = 0

f (x) = 6 x^{4} + 32 x^{3}

(3, - 1), (x, 8)

What is the extreme f(x)=-32x+36x^{1/2}+24 ?
The extreme f(x)=-32x+36x^{1/2}+24 is Maximum(0,24),Minimum(81/256 , 273/8)