## laurenn_marie7 2 years ago f(x)=2x^2+5x+2 ... find f(x+h)

1. camoJAX
2. Goten77

this means whereever there is a x in the original equasion that you substituate in x+h instead

3. Goten77

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4. laurenn_marie7

@Goten77 so I got to 2x^2+4xh+2h^2+5x+5h+2 ... is that as far as I can get?

5. Goten77

were the directions to expand it?

6. laurenn_marie7

No just to find the value

7. Goten77

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8. Goten77

oh then pluggign in x+h is all ya had to do

9. laurenn_marie7

oh.... so just 2(x+h)^2+5(x+h)+2 would be my answer?

10. laurenn_marie7

@Goten77

11. whpalmer4

That's the best you can do. You need to expand it, though. Usually this particular construct is in the context of $\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$and the subtraction of f(x) removes many of the terms and the h in the denominator cancels one or more, hopefully leaving you with a limit you can evaluate without too much pain and suffering. In the case of this problem, you would have $\lim_{h\rightarrow0}\frac{2x^2+4xh+2h^2+5x+5h+2-(2x^2+5x+2)}{h} =$$\lim_{h\rightarrow0}\frac{4xh+2h^2+5h}{h} =\lim_{h\rightarrow0}4x+2h+5 = 4x+5$

12. whpalmer4

And as you'll find out soon enough, $\frac{d}{dx}[2x^2+5x+2] = 4x+5$

13. laurenn_marie7

So confusing...so is it 4x+5 ? @whpalmer4

14. whpalmer4

Sorry, that was a preview of coming attractions :-) No your answer is $2x^2+4xh+2h^2+5x+5h+2$just like you asked about.

15. laurenn_marie7

Gotcha! Thank you so much! What about for -f(x)? @whpalmer4

16. whpalmer4

Unless there's a part of your problem statement that you didn't include or I didn't see, -f(x) isn't part of your problem, only my digression about computing derivatives...

17. laurenn_marie7

No no -f(x) of 2x^2+5x+2 @whpalmer4

18. whpalmer4

if $$f(x) = 2x^2+5x+2$$ then $$-f(x) = -(2x^2+5x+2) = -2x^2-5x-2$$, no?

19. whpalmer4

Not to be confused with $$f(-x) = 2(-x)^2+5(-x)+2 = 2x^2-5x+2$$