The Difference Quotient

[Euler]

The problem reads:

11b. Using a tolerance of h = 0.1, find the forward, backward, and symmetric difference quotients for f'(1).

If I could just understand the forward difference quotient, the backwards and symmetric one would make sense to me.

f(x) = x³ - x + 1

The forward difference quotient reads:
(f(x+h)-f(x))/h

When I plug in numbers, I always end up with 1 as my answer, when the book states it is 2.31

(f(x+h)-f(x))/h = (((1)³ - 1 + 1 + .1) - ((1)³ - 1 + 1)))/.1

This becomes .1/.1 which equals 1! Anyone know what exactly I am doing wrong?

 

[stapel]

I'm sorry, but I'm not following your work. How did you get that f(1.1) equalled "1 - 1 + 1 + 0.1"? Since f(x) has three terms, not four, shouldn't you have gotten three terms to simplify?

Please reply showing all of your steps, starting from "f(1.1) = (1.1)³ - (1.1) + 1". Thank you.

Eliz.

 

[Euler]

f(x+h)-f(x)
----------- This is the difference quotient
h

1³ - 1 + 1 + .1 - 1³ - 1 + 1
---------------------------------------------------------------
.1

The reason you see four terms 1³ - 1 + 1 + .1 is because it is f(x+h), not just f(x), at least on the left side.

Continuing on…

1 - 1 + 1 + .1 = 1.1 and 1 - 1 + 1 = 1. 1.1 - 1 = .1

Now, h = .1, so I am left with

.1
-- = 1
.1

The answer is 2.31

 

[Unco]

f(x + h) = f(1 + 0.1) = f(1.1) = (1.1)³ - (1.1) + 1 = 1.231

This doesn't seem to lead to 2.31, however.

 

[Euler]

f(x + h) = f(1 + 0.1) = f(1.1) = (1.1)³ - (1.1) + 1 = 1.231

This doesn't seem to lead to 2.31, however.

No no, that's perfect! You didn't subtract 1, and then divide by .1. I thought I had to plug in the equation first, but all I have to do is plug in the x-value and the tolerance. Thank you!

 

[Unco]

f(x + h) = f(1 + 0.1) = f(1.1) = (1.1)³ - (1.1) + 1 = 1.231

This doesn't seem to lead to 2.31, however.

No no, that's perfect! You didn't subtract 1, and then divide by .1. I thought I had to plug in the equation first, but all I have to do is plug in the x-value and the tolerance. Thank you!

Ah, I see. I must have added.