Chapter 2.3, Problem 85AYU

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.

91. $f\left(x\right)\text{ = 2}{x}^2\text{ - 3}x\text{ + 1}$

To determine

To find: The following values using the given function $f\left(x\right)=2x^2-3x+1$:

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 85AYU

a. $4x+2h-3$ 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 $1,1.2,1.02\text{ and}1$.

c. $y=1.02x-1.02$ is the equation for the secant line at $x=1$ with $h=0.01$.

Explanation of Solution

Given:

The function is given as $f\left(x\right)=2x^2-3x+1$.

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 $f\left(x\right)$,

$m_{sec}=\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{2{\left(x+h\right)}^2-3\left(x+h\right)+1-2x^2+3x-1}{h}=\frac{2x^2+2h^2+4xh-3x-3h-2x^2+3x}{h}=4x+2h-3$

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

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.5\right)-f\left(1\right)}{0.5}=4\left(1\right)+2\left(0.5\right)-3=1$

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

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.1\right)-f\left(1\right)}{0.1}=4\left(1\right)+2\left(0.1\right)-3=1.2$

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

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.01\right)-f\left(1\right)}{0.01}=4\left(1\right)+2\left(0.01\right)-3=1.02$

As $h$ approaches 0, $m_{sec}=4\left(1\right)-3=1$

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

$\left(1, f\left(1\right)\right)=\left(1, 0\right)$

$\left(1.01, f\left(1.01\right)\right)=\left(1.01, 0.0102\right)$

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

$\frac{y-0}{0.0102-0}=\frac{x-1}{1.01-1}$

$\frac{y}{0.0102}=\frac{x-1}{0.01}$

$y=1.02x-1.02$