Finding the Derivative: How to Solve for f'(x) Using the Limit Definition

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The discussion centers on finding the derivative of the function f(x) = 3x + 2 using the limit definition. The expression [f(a+h)-f(a)] / h simplifies to (3h + 4) / h, leading to confusion about how the book arrived at the derivative of 3. Participants clarify that the limit as h approaches 0 confirms the derivative is indeed 3, emphasizing that this specific case does not require taking the limit to reach the same result due to the constant nature of the function. The conversation highlights the importance of correctly applying the limit definition and recognizing potential mistakes in calculations. Ultimately, the derivative of the function is confirmed as 3, illustrating a fundamental concept in calculus.
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f(x)=3x+2 ; find [f(a+h)-f(a)] / h

So I got (3h+4)/h

But how did the book get 3? I tried everything. It must be an error.
 
Last edited:
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Assuming that
[f(a+h)-f(a)] / h
was actually
lim(h->0) [f(a+h)-f(a)] / h
the expression above is the first principle formula to get a derivative of a function
therefore, f'(x) = 3
Please don't rely on my solution entirely and please refer to others'

by the way, I am just curious that, how did you get (3h+4)/h?
 
CrossFit415 said:
f(x)=3x+2 ; find [f(a+h)-f(a)] / h

So I got (3h+4)/h

But how did the book get 3? I tried everything. It must be an error.

How did you start?
When I calculate it, I get 3.
 
I guess you didn't learn the differentiation rules yet? Or do you have to find the answer using that formula?

But yes, the answer would be 3.
 
Well...

[3(a+h)+2-3(a)+2] / h
=(3a+3h+2-3a+2) / h
=(3h+4) / h
 
conventionally, the following expression
lim(h->0) [f(a+h)-f(a)] / h
means the slope of the function at point a.

by the way, that's a creative thinking. Nice attempt.

In that expression
a is a certain 'point' at the function f(x) and h is a certain 'distance' from the point a.
You might already know that slope of a function is defined as
delta y / delta x
so let the slope between two points, a (a, f(a)) and p (a+h, f(a+h))
then the slope is defined as
(f(a+h)-f(a)) / (a+h - a)
which is [f(a+h)-f(a)] / h
now as h approaches to 0, you will get an instaneous slope of the function at the point a
 
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CrossFit415 said:
Well...

[3(a+h)+2-3(a)+2] / h
=(3a+3h+2-3a+2) / h
=(3h+4) / h

I see, you have left out a couple of parentheses, effectively mixing up additions and subtractions.

Try it with the following:

[(3(a+h)+2) - (3(a)+2)] / h

Cheers!
 
Yes^^
 
CrossFit415 said:
Well...

[3(a+h)+2-3(a)+2] / h
=(3a+3h+2-3a+2) / h
=(3h+4) / h

Your mistake is in the first line: f(a)=3a+2, thus -f(a)=-3a-2...
 
  • #10
I like Serena said:
I see, you have left out a couple of parentheses, effectively mixing up additions and subtractions.

Try it with the following:

[(3(a+h)+2) - (3(a)+2)] / h

Cheers!

Ahh careless mistakes again! Thank you!
 
  • #11
so it turns out to be that it is not a differntiation problem?
 
  • #12
It is a differentiation problem.
 
  • #13
Nothing tells us that it's a differentiation problem... He never took the limit of anything. For all we know, this could be an exercise in calculating with letters...
 
  • #14
I see..I was just under the impression that h approaches 0, and the derivative of that function is 3.
 
  • #15
This would typically be the set up in a math textbook just before introducing the derivative.
It starts with explaining slopes of functions over an interval h.

In this specific example the limit does not have to be taken to get the same result a derivative has, because the derivative is a constant function.
 
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