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# (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places.(b) Use calculus to find the exact maximum and minimum values.$f(x) = x \sqrt{x - x^2}$

## (b) $f(x)=x \sqrt{x-x^{2}} \Rightarrow f^{\prime}(x)=x \cdot \frac{1-2 x}{2 \sqrt{x-x^{2}}}+\sqrt{x-x^{2}}=\frac{\left(x-2 x^{2}\right)+\left(2 x-2 x^{2}\right)}{2 \sqrt{x-x^{2}}}=\frac{3 x-4 x^{2}}{2 \sqrt{x-x^{2}}}$So $f^{\prime}(x)=0 \Rightarrow 3 x-4 x^{2}=0 \Rightarrow x(3-4 x)=0 \Rightarrow x=0$ or $\frac{3}{4}$$f(0)=f(1)=0(\operatorname{minimum}),$ and $f\left(\frac{3}{4}\right)=\frac{3}{4} \sqrt{\frac{3}{4}-\left(\frac{3}{4}\right)^{2}}=\frac{3}{4} \sqrt{\frac{3}{16}}=\frac{3 \sqrt{3}}{16}(\operatorname{maximum})$

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the first thing we can do is graft this as we can see, the minimum value zero on the maximum value 0.32 approximately. Given the graphing calculator image. Now we're gonna move in, moving on to the calculus way, which means we can use the product rule to take the derivative F one g post if she won. Okay, If we set zero equals three X minus four X squared, we get axe times, three months for ox. Therefore, we get access for so our exes, 3/4. So now we know that half of zero zero off of 3/4 0.3 to 5 and then f of one is zero. The absolute maximum is three squirt of three over 16 which is the same thing as your 160.3 to 5 ever. Now in decimal form, who is easier to see which ones bigger and absolute minimum would obviously just be zero

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