SOLUTION: What is the difference quotient of the function f(x)=7x^2+6x+7? I know I must use the formula f(x)= [f(x+h)-f(x)]/h The first time I tried to solve it I ended up with [42x^+14xh+

Algebra ->  Test -> SOLUTION: What is the difference quotient of the function f(x)=7x^2+6x+7? I know I must use the formula f(x)= [f(x+h)-f(x)]/h The first time I tried to solve it I ended up with [42x^+14xh+      Log On


   

Question 551966: What is the difference quotient of the function f(x)=7x^2+6x+7?
I know I must use the formula f(x)= [f(x+h)-f(x)]/h
The first time I tried to solve it I ended up with [42x^+14xh+h^2+h]/h. Then I was told to use 7(x+h)^2 instead of (7x+h)^2, but I'm confused if I should do the same to 6x or not.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
First of all do NOT write f(x) = [f(x+h)-f(x)]/h because
f(x) = 7x² + 6x + 7.  Never write that something is equal to something
it is not equal to.  I'm sure you knew that f(x) was not equal to the
formula for the difference quotient, but what you wrote says just that.

You must write:

Difference quotient = %28f%28x%2Bh%29-f%28x%29%29%2Fh

f(x) = 7x² + 6x + 7

First find f(x+h)

f(x+h) = 7(x+h)² + 6(x+h) + 7 = 7(x²+2h+h²) + 6x+6h + 7 = 7x²+14hx+7h²+6x+6h+7

Then

difference quotient = %28f%28x%2Bh%29-f%28x%29%29%2Fh

Now we can substitute for both f(x+h) and f(x):

difference quotient = %28%287x%5E2%2B14hx%2B7h%5E2%2B6x%2B6h%2B7%29-%287x%5E2%2B6x%2B7%29%29%2Fh

difference quotient = %287x%5E2%2B14hx%2B7h%5E2%2B6x%2B6h%2B7-7x%5E2-6x-7%29%2Fh

We cancel the terms which cancel and we have

difference quotient = 

difference quotient = %2814hx%2B7h%5E2%2B6h%29%2Fh
 
factor h out of the numerator:

difference quotient = %28h%2814x%2B7h%2B6%29%29%2Fh

Cancel the h's

difference quotient = %28cross%28h%29%2814x%2B7h%2B6%29%29%2Fcross%28h%29

difference quotient = 14x+7h+6

Edwin