Question

if necessary, round results accurate to at least 2 decimal places. consider the function: (f(x)=15ln(x)-12sqrt{x}+10) on the interval (2,10) a. determine the absolute maximum of (f) on the given interval. if the extrema does not exist, enter \dne\ for both answers. (f) has an absolute maximum value of for both answers. b. determine the absolute minimum of (f) on the given interval. if the extrema does not exist, enter \dne\ for both answers. (f) has an absolute minimum value of which occurs at (x =) which occurs at (x =)

Explanation:

Step1: Find the derivative of $f(x)$

Use the derivative rules: $\frac{d}{dx}(\ln x)=\frac{1}{x}$, $\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}$.
$f'(x)=\frac{15}{x}-\frac{12}{2\sqrt{x}}=\frac{15}{x}-\frac{6}{\sqrt{x}}=\frac{15 - 6\sqrt{x}}{x}$

Step2: Find the critical points

Set $f'(x) = 0$, then $\frac{15 - 6\sqrt{x}}{x}=0$. Since $x
eq0$ in the domain of $f(x)$ on the open - interval $(2,10)$, we solve $15 - 6\sqrt{x}=0$.
$6\sqrt{x}=15$, $\sqrt{x}=\frac{15}{6}=\frac{5}{2}$, $x = (\frac{5}{2})^2=\frac{25}{4}=6.25$

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x = 2$, $x=6.25$, and $x = 10$.
$f(2)=15\ln(2)-12\sqrt{2}+10\approx15\times0.693 - 12\times1.414+10=10.395-16.968 + 10=3.427$
$f(6.25)=15\ln(6.25)-12\sqrt{6.25}+10=15\ln(6.25)-12\times2.5 + 10=15\ln(6.25)-20\approx15\times1.833 - 20=27.495-20 = 7.495$
$f(10)=15\ln(10)-12\sqrt{10}+10\approx15\times2.303-12\times3.162 + 10=34.545-37.944+10=6.601$

Answer:

a. $f$ has an absolute maximum value of $7.495$ which occurs at $x = 6.25$
b. $f$ has an absolute minimum value of $3.43$ (rounded to 2 decimal places) which occurs at $x = 2$