What is the extreme f(x)=(x^2-3)(x^{1/2}) ?

The extreme f(x)=(x^2-3)(x^{1/2}) is Maximum(0,0),Minimum((3/5)^{1/2},-(12*3^{1/4})/(5*5^{1/4)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x^2-3)(x^{1/2})

extreme points

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

Maximum (0, 0), Minimum ((\frac{3}{5})^{\frac{1}{2}}, - \frac{12 \cdot 3 ^{\frac{1}{4}}}{5 \cdot 5 ^{\frac{1}{4}}})

Solution steps

f^{'} (x) = \frac{( 5 ^{\frac{1}{2}} x + 3 ^{\frac{1}{2}} ) ( 5 ^{\frac{1}{2}} x - 3 ^{\frac{1}{2}} )}{2 x ^{\frac{1}{2}}}

Find intervals:

Decreasing

: 0 < x < (\frac{3}{5})^{\frac{1}{2}},

Increasing

: (\frac{3}{5})^{\frac{1}{2}} < x < \infty

Plug

x = 0

into

(x^{2} - 3) x^{\frac{1}{2}} : 0

Maximum (0, 0)

Plug

x = (\frac{3}{5})^{\frac{1}{2}}

into

(x^{2} - 3) x^{\frac{1}{2}} : - \frac{12 \cdot 3 ^{\frac{1}{4}}}{5 \cdot 5 ^{\frac{1}{4}}}

Minimum ((\frac{3}{5})^{\frac{1}{2}}, - \frac{12 \cdot 3 ^{\frac{1}{4}}}{5 \cdot 5 ^{\frac{1}{4}}})

Maximum (0, 0), Minimum ((\frac{3}{5})^{\frac{1}{2}}, - \frac{12 \cdot 3 ^{\frac{1}{4}}}{5 \cdot 5 ^{\frac{1}{4}}})

f (x) = sin (20 π t)

f (x) = - x^{2} + 9

x^{2} - 5 x + 4

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

y = 9 x - 6

What is the extreme f(x)=(x^2-3)(x^{1/2}) ?
The extreme f(x)=(x^2-3)(x^{1/2}) is Maximum(0,0),Minimum((3/5)^{1/2},-(12*3^{1/4})/(5*5^{1/4)})