Find f from this information f'(x)=ax+b, f(2)=0

  • Thread starter Thread starter ggcflo
  • Start date Start date
  • Tags Tags
    Information
Click For Summary
SUMMARY

The problem requires finding the function f given its derivative f'(x) = ax + b and the condition f(2) = 0. Integrating f'(x) yields f(x) = (a/2)x^2 + bx + C, where C is the constant of integration. Substituting x = 2 into the equation results in 2a + 2b + C = 0. However, the discussion concludes that there is insufficient information to determine a unique solution for f(x), as multiple functions can satisfy the given conditions.

PREREQUISITES
  • Understanding of calculus, specifically integration and differentiation
  • Familiarity with polynomial functions and their properties
  • Knowledge of the constant of integration in indefinite integrals
  • Ability to solve equations with multiple variables
NEXT STEPS
  • Study the concept of indefinite integrals and the role of the constant of integration
  • Explore polynomial function characteristics and their derivatives
  • Learn about boundary conditions and how they affect function determination
  • Investigate cases of underdetermined systems in calculus
USEFUL FOR

Students in calculus courses, mathematics educators, and anyone seeking to understand the implications of integration and boundary conditions in function determination.

ggcflo
Messages
6
Reaction score
0

Homework Statement



find f from this information f'(x)=ax+b, f(2)=0

Homework Equations


The Attempt at a Solution



if you integrate f'(x) treating a and b as constants you get f(x)= ax^2/2 + bx
and if you substitute f(2) you get 2a+2b. that's all i could get... don't know how you could use f(2)=0 in this problem :( help please.
 
Physics news on Phys.org
ggcflo said:
if you integrate f'(x) treating a and b as constants you get f(x)= ax^2/2 + bx

You forgot the constant of integration

If this is the whole problem, they don't give you enough information to fully solve for a single f(x) so there will be more than one right answer
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
8
Views
2K
Replies
4
Views
1K
Replies
6
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K