Find f(2+h), f((x+h) and f(x+h) - f(x)/h where h is equal to anything but zero for f(x) = x/x+1?

Question

I missed a calculus class and I am insure of how to solve this. Would I just set h in every equation to zero and try to solve for x?

Answer 1

As you have the function f(x) then you can easily find the functions f(2+h) and f(x+h) by just substituting the values 2+h and x+h in the place of x in the function f(x).

Illustration example, f(x) = 3x then f(2) = 3(2) = 6.

I am solving one for you and the other is also same as the first one.
f(2+h) = (2+h) / {(2+h) + 1} = (2+h) / (3+h)
In the same way you can get the uther one.

Now comming to the good part of your question.
In calculus, (According to limit concept)
Lt (h--0) { f(x+h) + f(x) } / h is defined as the differential of the given function.

The differential of f(x) is written as f^I (x) { You should keep a vertical dash after f and it is read as f dash x }

The third part of your question is to find f dash of x
F dash of x = { (x+1)(1) - (x)(1+0) } / (x+1)^2
= (x + 1 - x) / (x+1)^2
= 1 / (x+1)^2
So third answer is 1 / (x+1)^2
It is read as "one by x plus one wohle square"
square is only for x+1

The last question that you have asked is nice.
You should just set h in every equation to zero and try to solve for x because when you are evaluating the limits it is required to find the desired limit in the neighbourhood also . So this concept is used.

Hope you understand this and if you have any doubts then contact me.

Answer 2

Don't miss class! :)

f(2+h)=(2+h)/(2+h+1)

The second is more complicated and with the text format of the answers here, you'll need to be really careful with your parens:

(f(x+h)-f(x))/h=[(x+h)/(x+h+1)-x/(x+1)]/h=
[(x+h)(x+1)-x (x+h+1)]/[h (x+1)(x+h+1)]=
(x^2+x+xh+h-x^2-xh-x))/[h (x+1)(x+h+1)]=
h/[h (x+1)(x+h+1)]=
1/[(x+1)(x+h+1)]

So as h ->0, the limit is 1/(x+1)^2.

This is a classic calculation from differential calculus so be sure to carry your limits through each equation when you are requested to do so.

Good luck! That is a neat problem and the algebraic manipulation is a bit tricky so it is worth paying attention to the details.

Answer 3

f(2 + h) = F(x) = 2 + h / 2 + h + 1 = 2+ h / 3 + h

f(x + h) = F(x) = x + h / x + h + 1

F( x + h) - f(x)/h = f(x) = (x + h/ x +h +1) - (x / hx + h)

I think this is correct but I have been out of college for a couple of years

Answer 4

It's summer for God's sake. Not a chance in Hell am I solving that. good try though.

AND WHY DID YOU MISS CLASS? Naughty-naughty.

Answer 5

I believe you take the derivitive for f(x)..use the quotient rule..
and x cannot equal -1

Answer 6

1 lol

Answer 7

f(2+h)=(2+h)/(2+h+1)=(2+h)/(3+h)
f(x+h)=(x+h)/(x+h+1)


f(x+h)-f(x)/h=-(x+1-x)/(x+1)^2=-1/(x+1)^2

Answer 8

wow!!!!! I have no clue!