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Mathematics 38 Online
OpenStudy (anonymous):

Find the intervals on which the funtion g(x)=x sqrt(32-(x^2)) is increasing and decreasing. identify the function's local extreme values, if any, saying where they are taken on. Which, if any, of the extreme values are absolute?

OpenStudy (anonymous):

\[\begin{align*}g(x)&=x\sqrt{32-x^2}\\ g'(x)&=\left(32-x^2\right)^{1/2}+\frac{1}{2}x\left(32-x^2\right)^{-1/2}(-2x)\\ &=\left(32-x^2\right)^{1/2}-x^2\left(32-x^2\right)^{-1/2}\\ &=\left(32-x^2\right)^{-1/2}\left(32-x^2-x^2\right)\\ &=\frac{32-2x^2}{\sqrt{32-x^2}} \end{align*}\] Critical points: \[\frac{32-2x^2}{\sqrt{32-x^2}}=0~~\iff~~32-2x^2=0~~\iff~~x=\pm4\]

OpenStudy (anonymous):

The function's domain is all \(x\) with \(x^2\le32\), or \(-4\sqrt2\le x\le4\sqrt2\). So when you apply the derivative test, you set up the following intervals: \[(-4\sqrt2,-4),~(-4,4),~(4,4\sqrt2)\] Evaluate the derivative for some test point from each interval. For example, from the second interval, we could take \(x_t=0\), then \(g'(x_t)=\dfrac{32}{\sqrt{32}}\). The exact value is important. You're mainly interested in the sign of the value. In this case, \(g'(x_t)>0\), which means the graph of \(g(x)\) is increasing in this interval. Similarly, if \(g'(x_t)<0\), then \(g(x)\) is decreasing in the interval. Different patterns give different extrema: (1) increase then decrease, which indicates a maximum (2) decrease then increase, which indicates a minimum

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