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What is the easiest way to solve this?

2007-05-26 08:51:25 · 9 answers · asked by EboniDimplz 2 in Science & Mathematics Mathematics

9 answers

I agree, the easiest way is to take the derivative which is 2x.

however, to show your steps, plug (x+h) into the function. you'll get:
((x+h)^2+3-(x^2+3))/h
(x^2+2xh+h^2+3-x^2-3)/h
(2xh+h^2)/h
2x+h

2007-05-26 09:04:21 · answer #1 · answered by hrhbg 3 · 0 0

Since f(x) equals x^2+3 f(x+h) would equal (x+h)^2+3. The x in f(x) is similar to the x+h in f(x+h). The x+h is saying that the x in the function is shifted h units, so the h should only affect the x, so it would need to be (x+h)^2 + 3. So the final answer would have to be ((x+h)^2 + 3) - ((x^2+3)/h).

2007-05-26 16:03:58 · answer #2 · answered by schluect 2 · 1 0

f(x^2 + 3 +h) - f(x^2 + 3)/h

It would help if I knew what h means. (Is h what you are solving for? Or another function?)

2007-05-26 15:56:28 · answer #3 · answered by xo_cuddly_kitten_xo 4 · 0 0

Easiest way is to take the derivative, its 2x.

OR:

X^2+3+h-(x^2+3) all over h.

I'm sure you can take it from there.

2007-05-26 15:54:06 · answer #4 · answered by Anonymous · 0 0

you must substiture x in the first expression with x+h in the second expression:

f(x+h)= (x+h)^2+3
= x^2+h^2+2xh+3

and then subtract f(x)/h which is: (x^2+3)/h

all in all you get: x^2+h^2+2xh+3-x^2/h-3/h

2007-05-26 16:00:22 · answer #5 · answered by Anonymous · 0 0

f( x+h) = ( x + h )^2 + 3
= x^2 +2xh +h^2 + 3

f(x+h)-f(x) = (x^2 +2xh +h^2 + 3) - ( x^2 + 3 )
= x^2 +2xh +h^2 + 3 - x^2 -3
= 2xh + h^2

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

2007-05-26 17:55:43 · answer #6 · answered by muhamed a 4 · 0 0

everywhere u c f(x) just put in x^2+3

so

(x^2+3+h) - (x^2+3)/h

2007-05-26 15:55:36 · answer #7 · answered by SUPERMAN 4 · 0 0

If f(x)=x^2+3
f(x+h)=(x+h)^2+3=x^2+2hx+3
f(x+h)-f(x=2hx
f(x+h)-f(x)/h=2x
when hgoes to zero, this gives f'(x)
here f'(x)=2x.

2007-05-26 16:14:07 · answer #8 · answered by Anonymous · 1 0

f `(x) = ((x + h)² + 3 - (x² + 3)) / h
f ` (x) = (x² + 2hx + h² + 3 - x² - 3) / h
f ` (x) = (2hx + h²) / h
f `(x) = 2x + h
f `(x) = 2x as h-->0

2007-05-27 03:23:01 · answer #9 · answered by Como 7 · 0 0

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