Find the variable to make the function continuous

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To determine the value of k that makes the function j(x) continuous, the limits from both sides at x = 0 must be equal. The left limit is given by k cos(0) and the right limit by 10e(0) - k. Setting these two expressions equal allows for solving k. The discussion clarifies that continuity requires the function's value at x = 0 to match these limits. Participants confirm the approach of equating the limits to find k is correct.
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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