Chapter 3.FR3, Problem 1TU

To determine

To prove:

The equation of the secant line for the graph of $y=\sqrt{x}$ from $x=100$ to $x=121$ is $y=\frac{1}{21}x+\frac{110}{21}.$

Answer to Problem 1TU

Solution:

The slope is $\frac{1}{21}$ and proved that the equation of the secant line is $y=\frac{1}{21}x+\frac{110}{21}.$

Explanation of Solution

Calculation:

Given,

The slope of the secant line is the same as the average rate of change of $f\left(x\right)=\sqrt{x}$ from $x=100$ to $x=121$

$\text{Slope}=\text{Average rate of change}$

$=\frac{\sqrt{121}-\sqrt{100}}{121-100}$

$=\frac{11-10}{21}$

$\text{Slope}=\frac{1}{21}$

Thus, $y=\frac{1}{21}x+b$

When

$x=100$

$y=\sqrt{100}=10$

We can use these values to find the equation of the secant line

$y=\frac{1}{21}x+b$

$10=\frac{1}{21}\left(100\right)+b$

$10=\frac{100}{21}+b$

Add $\frac{-100}{21}$ on both sides,

$10-\frac{100}{21}=\frac{100}{21}+b-\frac{100}{21}$

$\frac{210-100}{21}=b$

$\frac{110}{21}=b$

Thus, the equation of the secant line is $y=\frac{1}{21}x+\frac{110}{21}.$

Hence, proved.

Final statement:

Therefore, the slope is $\frac{1}{21}$ and proved that the equation of the secant line is $y=\frac{1}{21}x+\frac{110}{21}.$