## anonymous one year ago For the function, find and simplify...will write equation below

1. anonymous

(A) $f(x+h)-f(x)$ and (B)$\frac{ f(h+h)-f(x) }{ h }$ $f(x)=9x^2-1$

2. butterflydreamer

We know our function is: f(x)=9x^2−1 For part (a), it wants you to find f( x + h) - f(x), So firstly you need to work out what "f (x+h)" equals to... To do this, you just sub (or "plug") in "x+h" into your original f(x) function... so replace the x's with (x+h)

3. butterflydreamer

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4. UsukiDoll

isn't it this function for b which is also the definition of the derivative in calculus I? $\frac{ f(x+h)-f(x) }{ h }$

5. anonymous

I honestly have no clue how to do this so I'm not sure

6. butterflydreamer

are you confused on how to find f(x+h) ?

7. anonymous

So I only have one example and here's how it goes...$F(x)=x^2+3x$ Step 1: $(x+h)^2+3(x+h)$ Now with this example that I opened the question with...I dont know how to set it up like this example

8. butterflydreamer

In that example, your function is f(x) = x^2 + 3x They found f(x + h) by REPLACING the "x" in "f(x) = x^2 + 3x" with "x+h" That's why we get: f(x + h) = (x + h) ^2 + 3 (x + h)

9. butterflydreamer

|dw:1434346176107:dw|

10. anonymous

So I think I'm confused because I don't know where the 9 should go because in the example I just showed you, x was 1

11. butterflydreamer

9 stays the same... Your entire function will remain exactly the same but you just replace the "x" with "x+h"

12. anonymous

So it should be (9x+h)^2-1(x+h)

13. butterflydreamer

no.. we ONLY replace the X. ORIGINAL function f(x) = 9x^2 -1 f(x+h) = 9(x+h)^2 - 1 we only replaced x with x+h We kept the 9 and the 1.

14. anonymous

oh ok. I think I can handle the rest now. I just didn't know how to set up my problem

15. butterflydreamer

sureee, good luck !