Alternative Difference Quotient

[avanm]

Hi there im trying to evaluate the alternative difference quotient f(x)-f(a)/x-a for the function 2-x-x^2. I'll attach my work I've tried to do I got -1-2(x+a). My friend did this with the original difference quotient of f(x+h)-f(x)/h and got -2h-4x-1.
Is what I did correct? (She used the difference quotient because her question required It but mine did not.)

 

[tkhunny]

Hi there im trying to evaluate the alternative difference quotient f(x)-f(a)/x-a for the function 2-x-x^2. I'll attach my work I've tried to do I got -1-2(x+a). My friend did this with the original difference quotient of f(x+h)-f(x)/h and got -2h-4x-1.
Is what I did correct? (She used the difference quotient because her question required It but mine did not.)

Have you considered that both might be incorrect?

Try to remember that f(x)-f(a)/x- a is NOT the same as [f(x)-f(a)]/[x- a]. Which is what you intended?

You must recognize that "a" and "h" are NOT independent. You tell me how they are related.

 

[avanm]

Have you considered that both might be incorrect?

Try to remember that f(x)-f(a)/x- a is NOT the same as [f(x)-f(a)]/[x- a]. Which is what you intended?

You must recognize that "a" and "h" are NOT independent. You tell me how they are related.

Hi yes [f(x)-f(a)]/[x- a] is what I intended! Sorry I got lazy there. are 'a' and 'h' basically the same thing and interchangeable? I saw an example of 1/x being evaluated by both difference quotients [f(x)-f(a)]/[x- a] and [f(x+h)-f(x)]/h and the first one was -1/xa and the other was -1/[x(h+x)] so thats why I said they're interchangeable. You should get the same answer just their used for different things?

Also, neither are correct?! Whats the right path?

Thank you!

 

[Dr.Peterson]

Hi there im trying to evaluate the alternative difference quotient f(x)-f(a)/x-a for the function 2-x-x^2. I'll attach my work I've tried to do I got -1-2(x+a). My friend did this with the original difference quotient of f(x+h)-f(x)/h and got -2h-4x-1.
Is what I did correct? (She used the difference quotient because her question required It but mine did not.)

These two forms are equivalent if you equate x in the second to a in the first, and h in the second to x-a in the first. That is, if in -1-2(x+a) you replace x with x+h and then a with x, you get -1-2(x+h+x) = -1-2(2x+h) = -1-4x-2h, which is what your friend got in answering what evidently was a different question. The question you answered is incomplete if it doesn't specify which form is required, as the two are different expressions.

But once you take the appropriate limit, you will get the same exact answer from both. When your x goes to a, it becomes f'(a) = -1-4a; when their h goes to zero, it becomes f'(x) = -4x-1. These are the same function.

 

[avanm]

These two forms are equivalent if you equate x in the second to a in the first, and h in the second to x-a in the first. That is, if in -1-2(x+a) you replace x with x+h and then a with x, you get -1-2(x+h+x) = -1-2(2x+h) = -1-4x-2h, which is what your friend got in answering what evidently was a different question. The question you answered is incomplete if it doesn't specify which form is required, as the two are different expressions.

But once you take the appropriate limit, you will get the same exact answer from both. When your x goes to a, it becomes f'(a) = -1-4a; when their h goes to zero, it becomes f'(x) = -4x-1. These are the same function.

Thank you this was a very helpful answer! My question required the alternative difference quotient and hers the x+h difference quotient. Thank you for clarifying these answers are the equivalent!