Given f(x)= x^2 - 2x +3, find f(x+h)?

👨🏻‍🏫 學術台 數學
2009-03-07 9:38 am
and if possible a second question needed to be answered

hence, express [ f(x+h)-f(x) ] / h in the simplest form

thank you
更新1:

ive replaced x with (x+h): (x+h)^2 - 2(x+h)+3 expanded: x^2 +2hx +h^2 -2x +2h +3 im not sure where to go from here? or is that the answer :/

回答 (11)

2009-03-07 9:47 am
✔ 最佳答案
REVISED with additional notes for you!

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

To check:
Pick two diff. variables for x and h. I used 3 for x and 4 for h. Plug these back into:

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

and you get 8.

Use the same x and h values plug it into h + 2x - 2 and you also get 8.

Tada!

What you did wrong: you're supposed to to do [f(x+h) - f(x)]/h
You only did f(x+h)
👍 0 👎 0
2009-03-07 9:44 am
1. f(x+h) = plug in (x+h) into (x) and then evaluate ( which is too long lol so you do yourself)

clue( (x+h)^2

2. i don't think you can evaluate it anymore since f(x+h) and F(x) is pretty much it own value so yo ucan't cancel it out.
👍 1 👎 0
2016-10-22 2:10 pm
Step #one million flow legally purchase a gun Step #2 Shoot the mother f**ker that made that equation..... --------------------------------------... Step #3 (elective)-in basic terms f**king Google it or grab a calculator..... O.o
👍 0 👎 0
2009-03-07 10:27 am
f(x) = x^2 - 2x + 3
f(x + h) = (x + h)^2 - 2(x + h) + 3
f(x + h) = (x + h)(x + h) - 2*x - 2*h + 3
f(x + h) = x^2 + hx + hx + h^2 - 2x - 2h + 3
f(x + h) = x^2 + 2hx + h^2 - 2x - 2h + 3
f(x + h) = x^2 - 2x + 2hx + h^2 - 2h + 3
👍 0 👎 0
2009-03-07 10:01 am
f(x+h)=x^2+2hx+h^2-2h-2x+3

[ f(x+h)-f(x) ] / h= [x^2+2hx+h^2-2h-2x+3-x^2+2x-3]/h
cancel out stuff
[2hx+h^2-2h]/h
factor h out of the top
h(h+2x-2)/h
cancel h's
h+2x-2 is answer
參考: MATH
👍 0 👎 0
2009-03-07 9:58 am
You did it right.
Just keep going, you'll get there.
It looks like your going to start working with limits.
A precurser to Calculas.
Good for you.
👍 0 👎 0
2009-03-07 9:52 am
Is should be:
f(x+h) = (x+h)^2 - 2(x+h) + 3
= x^2 + 2xh + h^2 - 2x -2h + 3
= x^2 + 2(h-1)x + h^2 - 2h + 3
[f(x+h)-f(x)]/h = [(x^2 + 2hx + h^2 - 2x - 2h + 3) - (x^2 - 2x + 3)]/h
= (h^2 + 2hx - 2h)/h
= h + 2x - 2
👍 0 👎 0
2009-03-07 9:57 am
f (x + h) = (x + h)² - 2 (x + h) + 3

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

f (x) = x ² - 2x + 3

f (x + h ) - f (x) = 2 h x + h ² - 2h

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

f ` (x) = 2x - 2
👍 0 👎 1
2009-03-07 9:54 am
f(x)= x^2 - 2x +3, find f(x+h)
f(x+h)=(x+h)^2-2(x+h)+3
=(x+h)(x+h)-2x-2h+3
.......=x^2+2hx-2x-2h+3 answer

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

=x^2+2hx-2x-2h+3-(x^2 - 2x +3)
------------------------------------------------
.................h

=x^2+2hx-2x-2h+3-x^2+2x-3
-------------------------------------------
.................h

= 2hx-2h
---------------
.....h

...2h(x-1)
=-----------
........h

=2(x-1) or 2x-2 answer//
👍 0 👎 1
2009-03-07 9:48 am
f(x+h) = (x+h)^2 - 2(x+h) +3

=x^2 + 2xh + h^2 - 2x - 2h +3

EDIT: - Sorry I did the subtraction in the wrong direction earlier. I've swapped it around now.

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

And all that can be simplified to 2xh +h^2 -2h

Divide that by h and you get 2x + h - 2
👍 0 👎 1


收錄日期: 2021-05-01 12:06:37
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