Problem

Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for the function $f(x)=5 x^{\frac{2}{3}}+2 x^{\frac{5}{3}}$ on the domain $[-2,1]$.

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Answer

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Answer

The absolute maximum of the function on the domain \([-2,1]\) is \(\boxed{7}\) at \(x = 1\), and the absolute minimum is \(\boxed{0}\) at \(x = 0\)

Steps

Step 1 :Find the derivative of the function \(f(x)=5 x^{\frac{2}{3}}+2 x^{\frac{5}{3}}\), which is \(f'(x) = \frac{10}{3} x^{-\frac{1}{3}} + \frac{10}{3} x^{\frac{2}{3}}\)

Step 2 :Set the derivative equal to zero and solve for \(x\), which gives \(x = -1\)

Step 3 :The derivative is undefined when \(x = 0\), so \(x = 0\) is also a critical point

Step 4 :Evaluate the function at the critical points and the endpoints of the domain: \(f(-2) = 5(-2)^{\frac{2}{3}} + 2(-2)^{\frac{5}{3}} = 5(4^{\frac{1}{3}}) - 2(32^{\frac{1}{3}}) = 20 - 2(2\sqrt[3]{4})\), \(f(-1) = 5(-1)^{\frac{2}{3}} + 2(-1)^{\frac{5}{3}} = 5 - 2 = 3\), \(f(0) = 5(0)^{\frac{2}{3}} + 2(0)^{\frac{5}{3}} = 0\), and \(f(1) = 5(1)^{\frac{2}{3}} + 2(1)^{\frac{5}{3}} = 5 + 2 = 7\)

Step 5 :The absolute maximum of the function on the domain \([-2,1]\) is \(\boxed{7}\) at \(x = 1\), and the absolute minimum is \(\boxed{0}\) at \(x = 0\)

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