2 more questions regarding limits with difference quotients. :( picklejuice.

The difference quotient is:

[ f (x+h) - (f-x) ] / h

the 1st function is as follows: f(x) = x^2 - 3x The answer is: 2x-3
This one was hard because of all the x's in the function, so I didn't know what to substitute and where.

the 2nd function is : f(x) = 1/ (x+2) The Answer is -1/ (x+2)^2

*Note: They are two seperate problems, and please do it step by step.

Thanks for helping me, in advance. Best regards! picklejuice...

(1) With f(x) = x^2 - 3x, we see that:
f(x + h) = (x + h)^2 - 3(x + h).

By the definition of the derivative (which is just the difference quotient but as h --> 0):
f'(x) = lim (h-->0) [f(x + h) - f(x)]/h
= lim (h-->0) [(x + h)^2 - 3(x + h) - (x^2 - 3x)]/h
= lim (h-->0) (x^2 + 2xh + h^2 - 3x - 3h - x^2 + 3x)/h, by expanding
= lim (h-->0) (2xh + h^2 - 3h)/h, by canceling other terms
= lim (h-->0) [h(2x + h - 3)]/h, by factoring out h
= lim (h-->0) (2x + h - 3), by canceling h
= 2x + 0 - 3
= 2x - 3, as required.

(2) With f(x) = 1/(x + 2), we have:
f(x + h) = 1/(x + h + 2).

So, we have:
f'(x) = lim (h-->0) [f(x + h) - f(x)]/h
= lim (h-->0) [1/(x + h + 2) - 1/(x + 2)]/h.

Multiplying the numerator and denominator by (x + 2)(x + h + 2) gives:
lim (h-->0) [1/(x + h + 2) - 1/(x + 2)]/h
= lim (h-->0) [x + 2 - (x + h + 2)]/[h(x + 2)(x + h + 2)]
= lim (h-->0) -h/[h(x + 2)(x + h + 2)]
= lim (h-->0) -1/[(x + 2)(x + h + 2)], by canceling h
= -1/[(x + 2)(x + 0 + 2)]
= -1/(x + 2)^2, as required.

I hope this helps!

[ f (x+h) - f(x) ] / h you typed it wrong !!!!
f(x) = x^2 - 3x
find f(x+h)=(x+h)^2-3(x+h)
=x^2+2xh+h^2-3x-3h
hence numerator
=x^2+2xh+h^2-3x-3h-(x^2-3x)
==x^2+2xh+h^2-3x-3h-x^2+3x
=2xh+h^2-3h
numerator/denominator
=2xh+h^2-3h / h
=2x-3
do the II one as a practice problem and succeed.