Find the variable to make the function continuous

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Homework Help Overview

The problem involves determining the value of k that makes the piecewise function j(x) continuous across its domain. The function is defined as j(x) = {k cos(x), x ≤ 0; 10ex − k, 0 < x.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to check the limits of both parts of the function and consider setting them equal to find k. There is also a question about the interpretation of the function's definition for x > 0.

Discussion Status

Some participants have offered guidance on checking the limits from both sides of zero and ensuring they coincide with the function's value at that point. There is an acknowledgment of initial misunderstandings regarding the problem's requirements.

Contextual Notes

There is a mention of potential confusion regarding the notation used in the function definition, specifically whether it is 10*e*x - k or 10*ex - k.

Painguy
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Homework Statement


Find k so that the following function is continuous on any interval.

j(x) = {k cos(x), x ≤ 0
{10ex − k, 0 < x

Homework Equations




The Attempt at a Solution


I originally thought i had to check if the limits of both parts of the functions existed, and if so to set them equal to each other, but then I reread the question and realized that it wasn't asking me if its continuous, but to make it continuous by finding the value of k. I figured that i might be able to to set the two functions equal to each other to see if I can get k, but I'm not sure if that's right or not.
 
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Painguy said:

Homework Statement


Find k so that the following function is continuous on any interval.

j(x) = {k cos(x), x ≤ 0
{10ex − k, 0 < x

Homework Equations




The Attempt at a Solution


I originally thought i had to check if the limits of both parts of the functions existed, and if so to set them equal to each other, but then I reread the question and realized that it wasn't asking me if its continuous, but to make it continuous by finding the value of k. I figured that i might be able to to set the two functions equal to each other to see if I can get k, but I'm not sure if that's right or not.

For x > 0 do you mean 10*e*x - k, or do you mean 10*ex - k?

RGV
 
Painguy said:

Homework Statement


Find k so that the following function is continuous on any interval.

j(x) = {k cos(x), x ≤ 0
{10ex − k, 0 < x

Homework Equations




The Attempt at a Solution


I originally thought i had to check if the limits of both parts of the functions existed, and if so to set them equal to each other, but then I reread the question and realized that it wasn't asking me if its continuous, but to make it continuous by finding the value of k. I figured that i might be able to to set the two functions equal to each other to see if I can get k, but I'm not sure if that's right or not.


Maybe to give a 100% answer, just state that right of zero and left of zero there are

no problems of continuity. Then , also, f is continuous if the value at a point coincides

with the right- and left- limits.
 
Ray Vickson said:
For x > 0 do you mean 10*e*x - k, or do you mean 10*ex - k?

RGV

I'm sorry i meant 10*ex - k
 
Painguy said:

Homework Statement


Find k so that the following function is continuous on any interval.

j(x) = {k cos(x), x ≤ 0
{10ex − k, 0 < x

Homework Equations



The Attempt at a Solution


I originally thought i had to check if the limits of both parts of the functions existed, and if so to set them equal to each other, but then I reread the question and realized that it wasn't asking me if its continuous, but to make it continuous by finding the value of k. I figured that i might be able to to set the two functions equal to each other to see if I can get k, but I'm not sure if that's right or not.
Chose k to make the following two limits equal to each other.

\lim_{x\to0^+}\,j(x)\,, this is where j(x) = 10ex − k .

\lim_{x\to0^-}\,j(x)\,, this is where j(x) = k cos(x) .

Then make sure that those limits equal j(0) (which of course, they will).
 
I see. That helps. That's what i intended to to at first, but I am not sure how i misinterpreted the question. Anyway thanks for all your help guys
 

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