in exercises 17 - 26, find the extreme values...
in exercises 17 - 26, find the extreme values of the function on the given interval. 21. f(x)=x + 3/x on 1,5.
Answer
# Explanation:
## Step1: Find the derivative
Differentiate $f(x)=x + \frac{3}{x}=x+3x^{-1}$ using the power - rule. The derivative $f'(x)=1 - 3x^{-2}=1-\frac{3}{x^{2}}$.
## Step2: Find critical points
Set $f'(x) = 0$. So, $1-\frac{3}{x^{2}}=0$. Then $\frac{3}{x^{2}}=1$, and $x^{2}=3$. Solving for $x$, we get $x=\pm\sqrt{3}$. But since our interval is $[1,5]$, we only consider $x = \sqrt{3}$.
## Step3: Evaluate the function at critical and end - points
Evaluate $f(x)$ at $x = 1$, $x=\sqrt{3}$, and $x = 5$.
When $x = 1$, $f(1)=1+\frac{3}{1}=4$.
When $x=\sqrt{3}$, $f(\sqrt{3})=\sqrt{3}+\frac{3}{\sqrt{3}}=\sqrt{3}+\sqrt{3}=2\sqrt{3}\approx3.46$.
When $x = 5$, $f(5)=5+\frac{3}{5}=\frac{25 + 3}{5}=\frac{28}{5}=5.6$.
# Answer:
The minimum value of $f(x)$ on $[1,5]$ is $2\sqrt{3}$ at $x=\sqrt{3}$, and the maximum value is $\frac{28}{5}$ at $x = 5$.