Question

question 19 - 1 point
find the absolute maximum and absolute minimum of the function f(x)=\frac{2x^{\frac{5}{2}}}{5}-\frac{2x^{\frac{3}{2}}}{3}-5 over the interval 0,4.
round your answer to two decimal places. 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 at x =
absolute minimum of at x =

Explanation:

Step1: Find the derivative of the function

First, rewrite the function $f(x)=2x^{\frac{5}{2}}- 2x^{\frac{3}{2}}-5$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=2\times\frac{5}{2}x^{\frac{5}{2}-1}-2\times\frac{3}{2}x^{\frac{3}{2}-1}=5x^{\frac{3}{2}}-3x^{\frac{1}{2}}=x^{\frac{1}{2}}(5x - 3)$.

Step2: Find the critical points

Set $f'(x) = 0$. Since $x^{\frac{1}{2}}(5x - 3)=0$, we have two cases: $x^{\frac{1}{2}}=0$ gives $x = 0$ and $5x-3=0$ gives $x=\frac{3}{5}=0.6$. Both $x = 0$ and $x = 0.6$ are in the interval $[0,4]$.

Step3: Evaluate the function at the critical points and endpoints

Evaluate $f(x)$ at $x = 0$, $x=0.6$ and $x = 4$.

• When $x = 0$, $f(0)=2\times0^{\frac{5}{2}}-2\times0^{\frac{3}{2}}-5=-5$.
• When $x = 0.6$, $f(0.6)=2\times(0.6)^{\frac{5}{2}}-2\times(0.6)^{\frac{3}{2}}-5$. Let $a=(0.6)^{\frac{1}{2}}\approx0.7746$. Then $f(0.6)=2a^{5}-2a^{3}-5=2\times(0.7746)^{5}-2\times(0.7746)^{3}-5\approx2\times0.278 - 2\times0.467-5=0.556 - 0.934 - 5=-5.378$.
• When $x = 4$, $f(4)=2\times4^{\frac{5}{2}}-2\times4^{\frac{3}{2}}-5=2\times32-2\times8 - 5=64 - 16-5=43$.

Answer:

Absolute maximum of $43$ at $x = 4$, Absolute minimum of $-5.38$ at $x = 0.6$