# What are the critical points of # f(x) = x*sqrt(8-x^2) #?

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To find the critical points of ( f(x) = x\sqrt{8-x^2} ), first, compute its derivative:

[ f'(x) = \sqrt{8-x^2} + x \left(\frac{-x}{\sqrt{8-x^2}}\right) ]

Set ( f'(x) ) equal to zero and solve for ( x ) to find the critical points. However, note that the function is not defined for ( x ) such that ( 8-x^2 < 0 ). So, we need to find where ( 8-x^2 ) is non-negative:

[ 8 - x^2 \geq 0 ]

Solve for ( x ):

[ x^2 \leq 8 ]

[ -\sqrt{8} \leq x \leq \sqrt{8} ]

So, the function is defined in the interval (-\sqrt{8} \leq x \leq \sqrt{8}).

Now, set ( f'(x) ) equal to zero:

[ \sqrt{8-x^2} - \frac{x^2}{\sqrt{8-x^2}} = 0 ]

[ \sqrt{8-x^2} = \frac{x^2}{\sqrt{8-x^2}} ]

[ (8-x^2) = x^2 ]

[ 8 = 2x^2 ]

[ x^2 = 4 ]

[ x = \pm 2 ]

Since ( f(x) ) is defined in the interval (-\sqrt{8} \leq x \leq \sqrt{8}), ( x = -2 ) and ( x = 2 ) are critical points within this interval.

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