Solve f'(c)*(b-a): Net Change at c

  • Thread starter Kibbel
  • Start date
In summary, the expression f'(c)*(b-a) represents the net change in y over the interval (b-a) and f'(c) is the slope of a tangent to the graph of f at point c. The values b and a are likely endpoints of the interval and this expression may be encountered in the context of the mean value theorem. Further context may be needed for clarification.
  • #1
Kibbel
27
0

Homework Statement


This is my own question. I am trying to understand what

f'(c)*(b-a) exactly means. I know its (deltaY/deltaX)*deltaX, which gives us deltaY, but does that mean its the net change in Y over the entire interval, or just at point c?
 
Physics news on Phys.org
  • #2


This doesn't inherently mean anything about y or x.

f'(c) is the derivative of f at c; that is:

[tex] \lim_{h\rightarrow 0} \frac{f(c+h)-f(c)}{h} [/tex]

f'(c) is the slope of a tangent to the graph of f at c.

Now b and a don't inherently mean anything either; they're probably endpoints of some interval. It looks like you're encountering this in the context of the mean value theorem or something like that. Maybe if you gave some more context, we could clarify this.
 

1. What is the meaning of "f'(c)" in the equation "Solve f'(c)*(b-a): Net Change at c"?

The notation "f'(c)" represents the derivative of the function f at the point c. It measures the rate of change of the function at that specific point.

2. How is the derivative of a function related to the net change at a specific point?

The derivative of a function is equal to the slope of the tangent line at that point. This slope represents the rate of change of the function at that point, which is equivalent to the net change.

3. What is the significance of multiplying the derivative by the difference between two points (b-a) in the equation?

Multiplying the derivative by the difference between two points (b-a) allows us to calculate the net change between those two points. This is because the derivative represents the rate of change, and multiplying it by the difference between two points gives us the total change over that interval.

4. Can this equation be used for any type of function, or only specific types?

This equation can be used for any differentiable function, meaning any function for which the derivative exists at every point. This includes most common functions such as polynomials, trigonometric functions, and exponential functions.

5. Is there a specific method or formula for solving this type of equation?

Yes, the general formula for solving "f'(c)*(b-a): Net Change at c" is to first find the derivative of the function at the given point c, then multiply it by the difference between the two points (b-a).

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
507
  • Calculus and Beyond Homework Help
Replies
1
Views
257
  • Calculus and Beyond Homework Help
Replies
9
Views
534
  • Calculus and Beyond Homework Help
Replies
3
Views
263
  • Calculus and Beyond Homework Help
Replies
8
Views
454
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
245
Back
Top