What is the extreme f(x)=sqrt(x)-\sqrt[3]{x} ?

The extreme f(x)=sqrt(x)-\sqrt[3]{x} is Maximum(0,0),Minimum(64/729 ,-4/27)

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extreme f(x)=sqrt(x)-\sqrt[3]{x}

extreme points

f (x) = x - 3 x

Maximum (0, 0), Minimum (\frac{64}{729}, - \frac{4}{27})

Solution steps

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

Find intervals:

Decreasing

: 0 < x < \frac{64}{729},

Increasing

: \frac{64}{729} < x < \infty

Plug

x = 0

into

x - 3 x : 0

Maximum (0, 0)

Plug

x = \frac{64}{729}

into

x - 3 x : - \frac{4}{27}

Minimum (\frac{64}{729}, - \frac{4}{27})

Maximum (0, 0), Minimum (\frac{64}{729}, - \frac{4}{27})

f (x) = ln (x^{2} + 7 x + 14), - 4 \leq x \leq 1

f (x) = \frac{x - 1}{x - 2}

f (x) = 2 x^{2} - 36 ln (x)

y = x^{2} - 8 x + 12

f (x) = x^{3} - x^{2} - x + 10

What is the extreme f(x)=sqrt(x)-\sqrt[3]{x} ?
The extreme f(x)=sqrt(x)-\sqrt[3]{x} is Maximum(0,0),Minimum(64/729 ,-4/27)