What is the extreme f(x)=((x^2-1)/((5-x^2)^{0.5)}) ?

The extreme f(x)=((x^2-1)/((5-x^2)^{0.5)}) is Minimum(0,-1/(5^{1/2)})

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extreme f(x)=((x^2-1)/((5-x^2)^{0.5)})

extreme points

f (x) = (\frac{x ^{2} - 1}{( 5 - x ^{2} ) ^{0.5}})

Minimum (0, - \frac{1}{5 ^{\frac{1}{2}}})

Solution steps

f^{'} (x) = \frac{- x ^{3} + 9 x}{( 5 - x ^{2} ) ^{\frac{3}{2}}}

Find intervals:

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < 5^{\frac{1}{2}}

Plug

x = 0

into

\frac{x ^{2} - 1}{( 5 - x ^{2} ) ^{0.5}} : - \frac{1}{5 ^{\frac{1}{2}}}

Minimum (0, - \frac{1}{5 ^{\frac{1}{2}}})

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 1

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 8

y = x^{\frac{4}{7}} (x^{2} - 2)

f (θ) = 2 θ - 3 sin (θ), 0 \leq θ \leq π

f (x) = x (x - 3)^{2} (x + 1)^{4}

What is the extreme f(x)=((x^2-1)/((5-x^2)^{0.5)}) ?
The extreme f(x)=((x^2-1)/((5-x^2)^{0.5)}) is Minimum(0,-1/(5^{1/2)})