Chapter 2.3, Problem 84AYU

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.

90. $f\left(x\right)\text{ = 2}{x}^2\text{ + }x$

To determine

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

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

Answer to Problem 84AYU

a. $4x+2h+1$ is the slope of the secant line of function $f\left(x\right)$.

Explanation of Solution

Given:

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

Calculation:

It requires the discussion of a secant line.

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 $f\left(x\right)$,

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

$=\frac{2x^2+2h^2+4xh +x+h-2x^2-x}{h}=4x+2h+1$

To determine

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

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

Answer to Problem 84AYU

b. $m_{sec}$ for $h=0.5, 0.1$, and $0.01$ at $x=1$ and $h$ approaches 0 are $6,5.2,5.02\text{ and }5$.

Explanation of Solution

Given:

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

Calculation:

It requires the discussion of a secant line.

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,

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

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

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.5\right)-f\left(1\right)}{0.5}=4\left(1\right)+2\left(0.5\right)+1=6$

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

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.1\right)-f\left(1\right)}{0.1}=4\left(1\right)+2\left(0.1\right)+1=5.2$

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

$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left(1+0.01\right)-f\left(1\right)}{0.01}=4\left(1\right)+2\left(0.01\right)+1=5.02$

As $h$ approaches 0, $m_{sec}=4\left(1\right)+1=5$.

To determine

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

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

Answer to Problem 84AYU

c. $y=5.02x-2.02$ is the equation for the secant line at $x=1$ with $h=0.01$.

Explanation of Solution

Given:

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

Calculation:

It requires the discussion of a secant line.

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,

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

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

$\left(1, f\left(1\right)\right)=\left(1, 3\right)$

$\left(1.01, f\left(0.01\right)\right)=\left(1.01, 3.0502\right)$

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

$\frac{y-3}{3.0502-3}=\frac{x-1}{1.01-1}$

$\frac{y-3}{0.0502}=\frac{x-1}{0.01}$

$y-3=5.02x-1$

$y=5.02x-2.02$

To determine

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

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

Answer to Problem 84AYU

d.

Explanation of Solution

Given:

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

Calculation:

It requires the discussion of a secant line.

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,

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

d.