precalculus difference quotient?

I have tried numerous times but cant seem to get this right.

find the difference quotient of f, that is find [f(x+h)-f(x)]/h where h cannot =0. Be sure to simplify.

f(x)=5x+3

I found a similar problem online and tried masking what to do but my answer, which was (5h+6)/h, didn’t seem to be right. Any help, or ideas? Thank you

Top 3 Answers

A few days ago

wayner122

Favorite Answer

First figure out what f(x+h) is

f(x+h) = 5(x+h) + 3

= 5x + 5h + 3

and you know that f(x) = 5x + 3

So [f(x+h) – f(x)] / h =

(5x + 5h + 3 – 5x – 3)/ h

=5h / h = 5.

I think you were doing it right but when you did f(x+h) – f(x) you got 5x + 5h – 5x + 3 instead of -3

4 years ago

f(x+h) = (x+h)^3 – one hundred twenty five = x^3 + 3x^2*h + 3x*h^2 + h^3 – one hundred twenty five f(x+h) – f(x) = x^3 + 3x^2h + 3xh^2 + h^3 – one hundred twenty five – (x^3 – one hundred twenty five) = 3x^2h + 3xh^2 + h^3 once you divide by ability of h, you get 3x^2 + 3xh + h^2 Jen

A few days ago

Maverick

this site shows you the work….much easier than for me to type it all in.

http://cis.stvincent.edu/carlsond/ma109/diffquot.html

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precalculus difference quotient?

precalculus difference quotient?

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