What is the extreme f(x)=x^{-5/7}(x+4) ?

The extreme f(x)=x^{-5/7}(x+4) is Minimum(10,(7*2^{2/7})/(5^{5/7)})

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extreme f(x)=x^{-5/7}(x+4)

extreme points

f (x) = x^{- \frac{5}{7}} (x + 4)

Minimum (10, \frac{7 \cdot 2 ^{\frac{2}{7}}}{5 ^{\frac{5}{7}}})

Solution steps

f^{'} (x) = \frac{2 ( x - 10 )}{7 x ^{\frac{12}{7}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 10,

Increasing

: 10 < x < \infty

Plug

x = 10

into

x^{- \frac{5}{7}} (x + 4) : \frac{7 \cdot 2 ^{\frac{2}{7}}}{5 ^{\frac{5}{7}}}

Minimum (10, \frac{7 \cdot 2 ^{\frac{2}{7}}}{5 ^{\frac{5}{7}}})

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

f (x, y) = 2 x^{3} + y^{2} - 24 x

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

f (x, y) = x^{2} + x y + y^{2} - 37 y + 456

f (x) = x^{3} + 3 x^{2} + 3 x - 4

What is the extreme f(x)=x^{-5/7}(x+4) ?
The extreme f(x)=x^{-5/7}(x+4) is Minimum(10,(7*2^{2/7})/(5^{5/7)})