Problem

- 1 hour
16
QUESTION 19 - 4 POINTS
Find the absolute maximum and absolute minimum of the function $f(x)=\frac{2 x^{\frac{3}{2}}}{5}-\frac{2 x^{\frac{3}{2}}}{3}+4$ over the interval $[0,3]$.
Enter an exact answer. If there is more than one value of $x$ in the interval at which the maximum or minimum occurs, you should use a comma to separate them.

Provide your answer below:
- Absolute maximum of $\square$ at $x=\square$
- Absolute minimum of $\square$ at $x=\square$

Answer

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Answer

Compare these values to find the absolute maximum and minimum: \(\boxed{\text{Absolute maximum of } 4 \text{ at } x=0}\) and \(\boxed{\text{Absolute minimum of } \frac{-12\sqrt{3}}{15}+4 \text{ at } x=3}\)

Steps

Step 1 :Simplify the function: \(f(x)=\frac{2 x^{\frac{3}{2}}}{5}-\frac{2 x^{\frac{3}{2}}}{3}+4\) to \(f(x)=\frac{-4 x^{\frac{3}{2}}}{15}+4\)

Step 2 :Find the derivative of the function: \(f'(x)=\frac{-6 x^{\frac{1}{2}}}{15}\)

Step 3 :Set the derivative equal to zero to find the critical point: \(\frac{-6 x^{\frac{1}{2}}}{15}=0\) gives \(x=0\)

Step 4 :Evaluate the function at the endpoints of the interval and at the critical point: \(f(0)=4\) and \(f(3)=\frac{-12\sqrt{3}}{15}+4\)

Step 5 :Compare these values to find the absolute maximum and minimum: \(\boxed{\text{Absolute maximum of } 4 \text{ at } x=0}\) and \(\boxed{\text{Absolute minimum of } \frac{-12\sqrt{3}}{15}+4 \text{ at } x=3}\)

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