consider the following function. g(x) = 1/20 x^2 - 25√x on 0, 36 find the derivative of the function. g(x) =…

Explanation:

Step1: Find the derivative

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $y=\frac{1}{20}x^{2}-25x^{\frac{1}{2}}$, the derivative $g'(x)=\frac{1}{20}\times2x-25\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{10}x-\frac{25}{2\sqrt{x}}$.

Step2: Find the critical numbers

Set $g'(x) = 0$. So $\frac{1}{10}x-\frac{25}{2\sqrt{x}}=0$. Multiply through by $10\sqrt{x}$ to get $x^{\frac{3}{2}} - 125=0$. Then $x^{\frac{3}{2}}=125$, and $x = 25$. Also, $g'(x)$ is undefined at $x = 0$. So the critical numbers are $x=0,25$.

Step3: Evaluate the function at critical numbers and endpoints

Evaluate $g(x)$ at $x = 0$, $g(0)=\frac{1}{20}(0)^{2}-25\sqrt{0}=0$. Evaluate $g(x)$ at $x = 25$, $g(25)=\frac{1}{20}(25)^{2}-25\sqrt{25}=\frac{625}{20}-125=\frac{625 - 2500}{20}=-\frac{1875}{20}=-93.75$. Evaluate $g(x)$ at $x = 36$, $g(36)=\frac{1}{20}(36)^{2}-25\sqrt{36}=\frac{1296}{20}-150=\frac{1296 - 3000}{20}=-\frac{1704}{20}=-85.2$.

Answer:

$g'(x)=\frac{1}{10}x-\frac{25}{2\sqrt{x}}$ $x = 0,25$ maximum: $0$ minimum: $- 93.75$