What is the extreme f(x)=x^{1/2}(x-4) ?

The extreme f(x)=x^{1/2}(x-4) is Maximum(0,0),Minimum(4/3 ,-(16)/(3*3^{1/2)})

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extreme f(x)=x^{1/2}(x-4)

extreme points

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

Maximum (0, 0), Minimum (\frac{4}{3}, - \frac{16}{3 \cdot 3 ^{\frac{1}{2}}})

Solution steps

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

Find intervals:

Decreasing

: 0 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

x^{\frac{1}{2}} (x - 4) : 0

Maximum (0, 0)

Plug

x = \frac{4}{3}

into

x^{\frac{1}{2}} (x - 4) : - \frac{16}{3 \cdot 3 ^{\frac{1}{2}}}

Minimum (\frac{4}{3}, - \frac{16}{3 \cdot 3 ^{\frac{1}{2}}})

Maximum (0, 0), Minimum (\frac{4}{3}, - \frac{16}{3 \cdot 3 ^{\frac{1}{2}}})

\frac{2 x + 1}{x - 3}

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

f (x) = 7 x - x^{2}

f (x) = ln (4 - ln (x))

f (x) = x^{2} - y^{2} x - x y

What is the extreme f(x)=x^{1/2}(x-4) ?
The extreme f(x)=x^{1/2}(x-4) is Maximum(0,0),Minimum(4/3 ,-(16)/(3*3^{1/2)})