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mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Difference Quotient of f

Michael Sullivan introduces the calculus idea of the difference quotient of f at the end of section 3.3 in chapter 3.

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

m_sec = [f(x + h) - f(x)]/[(x + h) - x] which leads to [f(x + h) - f(x)]/h, where h cannot = 0.

A. Express the slope of the secant line for the function f(x) = 2x + 5 in terms of x and h. Be sure to simplify.

Note: m_sec = slope of the secant line.

sec_m = [2(x + h) + 5 - (2x + 5)]/h

My answer is 2.

You say?

B. Find m_sec for h = 0.5, 0 1, and 0.01 at x = 1. What value does m_sec approach as h approaches 0?

After working it out on paper, I get m_sec = 2. I also get 2 when h = 0.1 and h = 0.01. I conclude that m_sec tends to 2 as h tends to 0.

C. Is this what we call the limit in calculus?

D. Find the equation of the secant line at x = 1 with h = 0.01.

I need help with part D.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Difference Quotient of f

A, B, and C all correct.

D.sec_m = [2(x + h) + 5 - (2x + 5)]/h

Find the equation of the secant line at x = 1 with h = 0.01.

Sub in those values

sec_m = [2 times 1.01 + 5 - (2 + 5)]/0.01 = 0.02/0.01 = 2

So the line has gradient 2.

function is y = 2x + 5 so if x = 1, y = 7

So we have a line that has gradient 2 and goes through (1,7)

Because we're dealing with a straight line that is going to give y = 2x + 5 again.

Not sure why this question is even being asked.

I'd have chosen a function that isn't a line, eg y = x^2

Maybe that's the next question?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Difference Quotient of f

A, B, and C all correct.

D.sec_m = [2(x + h) + 5 - (2x + 5)]/h

Find the equation of the secant line at x = 1 with h = 0.01.

Sub in those values

sec_m = [2 times 1.01 + 5 - (2 + 5)]/0.01 = 0.02/0.01 = 2

So the line has gradient 2.

function is y = 2x + 5 so if x = 1, y = 7

So we have a line that has gradient 2 and goes through (1,7)

Because we're dealing with a straight line that is going to give y = 2x + 5 again.

Not sure why this question is even being asked.

I'd have chosen a function that isn't a line, eg y = x^2

Maybe that's the next question?

Bob

Yes, part D is weird.