Chapter 2.3, Problem 88AYU

Problems 87-94 require the following discussion of a secant line. The slope of the secant line containing the two points $(x,f(x))$ and $(x\text{ + }h,f(x\text{ + }h))$ on the graph of a function $y\text{ = }f\left(x\right)$ may be given as

In calculus, this expression is called the difference quotient of f

(a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer.

(b) Find m_sec for $h=0.5$, 0.1, and 0.01 at $x\text{ = 1}$. What value does m_sec approach as h approaches 0?

(c) Find an equation for the secant line at $x\text{ = 1}$ with $h\text{ = 0}\text{.01}$.

(d) Use a graphing utility to graph f and the secant line found in part (c) in the same viewing window.

94. $f\left(x\right)\text{ = }\frac{1}{x^2}$

To determine

To find: The following values using the given function $f\left(x\right)= \frac{1}{x^2}$:

a. Express and simplify the slope of the secant line of function in terms of $x$ and $h$.

b. $m_{sec}$ for $h=0.5, 0.1$, and $0.01$ at $x=1$.

c. An equation for the secant line at $x=1$ with $h=0.01$.

d. Graph the function $f$ and the secant line found in part $\left(c\right)$.

Answer to Problem 88AYU

a. $-\frac{h+2x}{x^2{\left(x+h\right)}^2}$ is the slope of the secant line of function $f\left(x\right)$.

b. $m_{sec}$ for $h=0.5, 0.1$, and $0.01$ at $x=1$ and $h$ approaches 0 are $-1.11,-1.735,-1.97\text{ and}-2$.

c. $\underset{\_}{y=-2x+3}$ is the equation for the secant line at $x=1$ with $h=0.01$.

Explanation of Solution

Given:

The function is given as $f\left(x\right)=\frac{1}{x^2}$.

It requires the discussion of a secant line.

Formula used:

The slope of the secant line containing the two points $\left(x, f\left(x\right)\right)$ and $\left(x+h, f\left(x+h\right)\right)$ on the graph of a function $y=f\left(x\right)$ may be given as

$m_{sec}=\frac{f\left(x+h\right)-f\left(x\right)}{h}$

a. Slope of the secant line of function $f\left(x\right)$,

$m_{sec}=\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\frac{1}{{\left(x+h\right)}^2}-\frac{1}{x^2}}{h}=\frac{\frac{x^2-x^2-h^2-2xh}{x^2{\left(x+h\right)}^2}}{h}= -\frac{h+2x}{x^2{\left(x+h\right)}^2}$

b. $m_{sec}$ for $h=0.5$ and $x=1$.

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.5\right)-f\left(1\right)}{0.5}=-\frac{0.5+2\left(1\right)}{{\left(1\right)}^2{\left(1+0.5\right)}^2}=-1.11$

$m_{sec}$ for $h=0.1$ and $x=1.1$

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.1\right)-f\left(1\right)}{0.1}=-\frac{0.1+2\left(1\right)}{{\left(1\right)}^2{\left(1+0.1\right)}^2}=-1.735$

$m_{sec}$ for $h=0.01$ and $x=1$.

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.01\right)-f\left(1\right)}{0.01}=-\frac{0.01+2\left(1\right)}{{\left(1\right)}^2{\left(1+0.01\right)}^2}=-1.97$

As $h$ approaches 0, $m_{sec}=-\frac{0+2\left(1\right)}{{\left(1\right)}^2{\left(1+0\right)}^2}=-2$

c. An equation for the secant line at $x=1$ with $h=0.01$.

The points are $\left(x, f\left(x\right)\right)$ and $\left(x+h, f\left(x+h\right)\right)=\left(1, f\left(1\right)\right)$ and $\left(1.01, f\left(1.01\right)\right)$

$\left(1, f\left(1\right)\right)=\left(1, 1\right)$

$\left(1.01, f\left(1.01\right)\right)=\left(1.01, 0.98\right)$

Equation of a straight line with two points $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$\frac{y-1}{0.98-1}=\frac{x-1}{1.01-1}$

$\frac{y-1}{-0.02}=\frac{x-1}{0.01}$

$y=-2x+2+1$

$y=-2x+3$