consider the function $f(x)=6sqrt{x}+3$ on the interval $1,10$. find the average or mean slope of the…

consider the function $f(x)=6sqrt{x}+3$ on the interval $1,10$. find the average or mean slope of the function on this interval.\n\nby the mean value theorem, we know there exists a $c$ in the open interval $(1,10)$ such that $f(c)$ is equal to this mean slope. for this problem, there is only one $c$ that works. find it.

consider the function $f(x)=6sqrt{x}+3$ on the interval $1,10$. find the average or mean slope of the function on this interval.\n\nby the mean value theorem, we know there exists a $c$ in the open interval $(1,10)$ such that $f(c)$ is equal to this mean slope. for this problem, there is only one $c$ that works. find it.

Answer

Explanation:

Step1: Recall the formula for average slope

The average slope of a function $y = f(x)$ on the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 1$, $b=10$, and $f(x)=6\sqrt{x}+3$.

Step2: Calculate $f(10)$ and $f(1)$

$f(10)=6\sqrt{10}+3$, $f(1)=6\sqrt{1}+3=6 + 3=9$.

Step3: Compute the average slope

The average slope $m=\frac{f(10)-f(1)}{10 - 1}=\frac{6\sqrt{10}+3 - 9}{9}=\frac{6\sqrt{10}-6}{9}=\frac{2\sqrt{10}-2}{3}$.

Step4: Find the derivative of $f(x)$

$f(x)=6x^{\frac{1}{2}}+3$, so $f'(x)=6\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{3}{\sqrt{x}}$.

Step5: Set $f'(c)$ equal to the average slope

We set $\frac{3}{\sqrt{c}}=\frac{2\sqrt{10}-2}{3}$. Cross - multiply to get $9=(2\sqrt{10}-2)\sqrt{c}$. Then $\sqrt{c}=\frac{9}{2\sqrt{10}-2}$. Rationalize the denominator: $\sqrt{c}=\frac{9(2\sqrt{10}+2)}{(2\sqrt{10}-2)(2\sqrt{10}+2)}=\frac{18\sqrt{10}+18}{40 - 4}=\frac{18\sqrt{10}+18}{36}=\frac{\sqrt{10}+1}{2}$. Square both sides to find $c = (\frac{\sqrt{10}+1}{2})^2=\frac{10 + 2\sqrt{10}+1}{4}=\frac{11 + 2\sqrt{10}}{4}$.

Answer:

Average slope: $\frac{2\sqrt{10}-2}{3}$ $c$ value: $\frac{11 + 2\sqrt{10}}{4}$