What Are the Values of c for Continuity at x=5?

Click For Summary
SUMMARY

The discussion focuses on determining the values of c that ensure the continuity of the piecewise function defined as f(x) = x^2 - c^2 for x < 5 and f(x) = cx + 11 for x ≥ 5. To find these values, participants emphasize the need to equate the limits of f(x) as x approaches 5 from both sides and ensure that f(5) matches these limits. The equation derived from this process is (5)^2 - c^2 = 5c - 11, which must be solved to find the specific values of c.

PREREQUISITES
  • Understanding of piecewise functions
  • Knowledge of limits in calculus
  • Ability to solve algebraic equations
  • Familiarity with continuity concepts in mathematics
NEXT STEPS
  • Study the concept of limits in calculus, focusing on one-sided limits
  • Learn how to solve piecewise functions and their continuity conditions
  • Explore algebraic techniques for solving quadratic equations
  • Review the implications of continuity in real-valued functions
USEFUL FOR

Students studying calculus, particularly those focusing on continuity and limits, as well as educators looking for examples of piecewise function analysis.

togame
Messages
18
Reaction score
0

Homework Statement


[tex]f(x) = x^2 - c^2 \mbox{ if } x < 5[/tex]
[tex]f(x) = cx+11 \mbox{ if } x \geq 5[/tex]

Find the two values of c for which the function would be continuous.


Homework Equations





The Attempt at a Solution


I set these two equations equal to each other, plug in the value 5 since that is the point at which these equations would meet, then solve for c? I'm not sure if I'm missing a step in the algebra or something else, but I seem to be unable to get the correct answer.
 
Last edited:
Physics news on Phys.org
togame said:

Homework Statement


[tex]f(x) = x^2 - c^2 \mbox{ if } x < 5[/tex]
[tex]f(x) = cx+11 \mbox{ if } x \geq 5[/tex]

Find the two values of c for which the function would be continuous.

Homework Equations



The Attempt at a Solution


I set these two equations equal to each other, plug in the value 5 since that is the point at which these equations would meet, then solve for c? I'm not sure if I'm missing a step in the algebra or something else, but I seem to be unable to get the correct answer.
Technically: You should find the limit of f(x) as x approaches 5 from the left, and then from the right and set the limits equal to each other, solving for c. Also, make sure that f(5) is the same as those limits.

Of course, when you do that, you do get [itex](5)^2-c^2=5c-11\,.[/itex]
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
22
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
Replies
8
Views
2K