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Chapter 4: Problem 30

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=x^{2} \sqrt{9-x^{2}} \text { on }(-3,3)$$

Short Answer

Expert verified
Answer: The function is increasing on the intervals $$(-3, -1)$$ and $$(0, 3)$$ and decreasing on the interval $$(-1, 0)$$

Step by step solution

01

Find the derivative of the function

Using the Product Rule and Chain Rule, we'll find the derivative of $$f(x)=x^{2}\sqrt{9-x^{2}}.$$ The Product Rule states that if $$f(x) = g(x)h(x),$$ then $$f'(x) = g'(x)h(x) + g(x)h'(x).$$ So, $$ g(x)=x^{2} \text{ and } h(x)=\sqrt{9-x^{2}}. $$ First, we find the derivative of each function. For $$g(x), g'(x)=\frac{d}{dx}(x^{2})=2x.$$ For $$h(x),$$ we rewrite it as $$h(x)=(9-x^{2})^{\frac{1}{2}}$$ and apply the Chain Rule: $$h'(x) = \frac{1}{2}(9-x^{2})^{-\frac{1}{2}}(-2x).$$ Now, apply the Product Rule: $$f'(x) = g'(x)h(x) + g(x)h'(x) \Rightarrow f'(x) = (2x)(\sqrt{9-x^{2}}) + (x^{2})(\frac{1}{2}(9-x^{2})^{-\frac{1}{2}}(-2x)).$$
02

Find the critical points

To find the critical points, set $$f'(x) = 0$$ and solve for x: $$0 = (2x)(\sqrt{9-x^{2}}) + (x^{2})(\frac{1}{2}(9-x^{2})^{-\frac{1}{2}}(-2x)).$$ First, check if $$2x = 0$$ so x = 0 is a critical point. Next, isolate the square root: $$\sqrt{9-x^{2}}=-\frac{x^{3}}{2},$$ square both sides: $$(9-x^{2})=\frac{x^{6}}{4},$$ and solve for x. The critical points are x = -1 and x = 1.
03

Determine the intervals of increase and decrease

Using the critical points and the endpoints of the given interval \((-3,3)\), we'll test the sign of $$f'(x)$$ in each subinterval: - For x in \((-3, -1)\), let's test x = -2: $$f'(-2) = (2(-2))(\sqrt{9-(-2)^{2}})$$ is positive. - For x in \((-1,0)\), let's test x = -0.5: $$f'(-0.5) = (2(-0.5))(\sqrt{9-(-0.5)^{2}})$$ is negative. - For x in \((0,1)\), let's test x = 0.5: $$f'(0.5) = (2(0.5))(\sqrt{9-(0.5)^{2}})$$ is positive. - For x in \((1,3)\), let's test x = 2: $$f'(2) = (2(2))(\sqrt{9-(2)^{2}})$$ is positive. Thus, the intervals of the increase are $$x \in (-3,-1)$$ and $$x \in (0,3)$$ while the interval of the decrease is $$x \in (-1,0).$$

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