For f(x) = x^2 + 9x, find [f(x+h)-f(x)]/h, [f(w)-f(x)]/(w-z)

[dominator236]

Help University integral calc 126

For the function f(x)= x^2+9x simplify each expression as much as possible

1. [f(x+h)-f(x)]/h, h cannot equal 0

2. [f(w)-f(x)]/(w-z), x cannot equal w

 

[stapel]

Help University integral calc 126

Since you are in an integral calculus course, if seems safe to assume that you have taken algebra, trig, and differential calculus.

For the function f(x)= x^2+9x simplify each expression as much as possible

1. [f(x+h)-f(x)]/h, h cannot equal 0

2. [f(w)-f(x)]/(w-z), x cannot equal w

Each of these is an evaluation/simplification exercise, like what you did back in algebra. So please re-read the "Read Before Posting" message, and reply with the requested information, starting with how you plugged the given arguments into the specified function.

Thank you!

 

[Steven G]

What does f(x+h) equal? What does f(x) equal. What do you get when you subtract them, ie what does f(x+h) - f(x) equal.
What does the 126 mean in integral calculus 126?

 

[Steven G]

Since you are in an integral calculus course, if seems safe to assume that you have taken algebra, trig, and differential calculus.

taken \(\displaystyle \neq\) learned

 

[dominator236]

What does f(x+h) equal? What does f(x) equal. What do you get when you subtract them, ie what does f(x+h) - f(x) equal.
What does the 126 mean in integral calculus 126?

So first the 126 was the university course code I thought maybe someone would recognize it so my mistake. Second, I figured it out finally but basically, I was doing (x^2+9x+h) - (x^2+9x). Instead of applying the ^2 to all of the terms in the bracket I still don't really get why we do that but thanks for your time!

 

[stapel]

So first the 126 was the university course code I thought maybe someone would recognize it so my mistake.

There are thousands of colleges and universities. They do not all share the exact same curriculum or course catalog.

Second, I figured it out finally but basically, I was doing (x^2+9x+h) - (x^2+9x).

It is most definitely *not* "doing f(x) + h - f(x)", which is what is meant by the above.

Instead of applying the ^2 to all of the terms in the bracket I still don't really get why we do that but thanks for your time!

I'm sorry, but I don't know what this means.

Try going back to what you saw back in algebra. You are given [imath]f(x) = x^2 + 9x[/imath]. You are told to find [imath]f(x+h)[/imath]. So find it:

[imath]\qquad f(x + h) = (x + h)^2 + 9(x + h)[/imath]

[imath]\qquad \qquad = x^2 + 2xh + h^2 + 9x + 9h[/imath]

Now subtract [imath]f(x)[/imath] from this. Then divide the result by [imath]h[/imath]. What remains when you are done?