Calculus: Preperation for Limits

[KidInkFan]

f(x): 2/x-1

Find an expression for f(x+h)-f(x)/h and simplify where h cannot equal 0.

I tried putting in the function of f(x) into the expression above and I got: ((2/x+h-1)-(2/x-1))/h but I'm uncertain of what to do afterwards i though if i got a common denominator that would help. I added common factors and such to get an answer of: -2h^2/(x+h-1)(x-1) but i dont think im doing it right.

Also, how would you simplify the following expression:

f(x): (squareroot x^4+3x^2)/x where x is negative. I don't understand what it means by where x is negative :S

Any help towards this problem is much appreciated!! Thank you!!!

 

[MarkFL]

Is the function:

\(\displaystyle f(x)=\dfrac{2}{x}-1\)

or

\(\displaystyle f(x)=\dfrac{2}{x-1}\) ?

It appears the second choice is the case, but it is important to use bracketing symbols to make things unambiguous.

Also, just a slight quibble...you are actually preparing for differentiation by using the definition of the derivative by first principles, which involves a limit.

 

[JeffM]

f(x): 2/x-1

Find an expression for f(x+h)-f(x)/h and simplify where h cannot equal 0.

I tried putting in the function of f(x) into the expression above and I got: ((2/x+h-1)-(2/x-1))/h but I'm uncertain of what to do afterwards i though if i got a common denominator that would help. I added common factors and such to get an answer of: -2h^2/(x+h-1)(x-1) but i dont think im doing it right.

Also, how would you simplify the following expression:

f(x): (squareroot x^4+3x^2)/x where x is negative. I don't understand what it means by where x is negative :S

Any help towards this problem is much appreciated!! Thank you!!!

\(\displaystyle f(x) = \dfrac{2}{x - 1}.\) As Mark said, please pay attention to proper bracketing. We are not mind readers.

Your idea of common denominators is absolutely correct, but then your algebra went off the rails. Because you did not show your work, I have no idea where.

\(\displaystyle h \ne 0 \implies \dfrac{f(x + h) - f(x)}{h} = \dfrac{\dfrac{2}{x + h + 1} - \dfrac{2}{x + 1}}{h} = \dfrac{2}{h(x + h + 1)} - \dfrac{2}{h(x + 1)} = \dfrac{2(x + 1) - 2(x + h + 1)}{h(x + 1)(x + h + 1)} = \)

\(\displaystyle \dfrac{2x + 2 - 2x - 2h - 2}{h(x + 1)\{(x + 1) + h\}} = \dfrac{-2h}{h\{(x + 1)^2 + h(x + 1)} = \dfrac{-2}{(x + 1)^2 + h(x + 1)}.\)

As for your second problem, you are overthinking it.

\(\displaystyle x < 0 \implies \dfrac{\sqrt{x^4 + 3x^2}}{x} = \dfrac{\sqrt{x^4 + 3x^2}}{- \sqrt{x^2}}.\) Proceed.

 

[KidInkFan]

Thank-you very much for your help!! I will keep in mind about putting brackets around a function when I have a question!