Let I be a nonempty subset of R, let f:I->R be a function. Let c be a member of

ID: 2944050 • Letter: L

Question

Let I be a nonempty subset of R, let f:I->R be a function. Let c be a member of the real numbers and define the function g:I->R by g(x)=f(cx), for x in I. Prove the following implication: If f is continious, then g is continuous. Is the converse true?

(I think that I know how to show the implication. I think you have to talk about 2 cases. case 1 cx is in I and then g(x) is continious because f(x) is continiouse at every point in I. case 2 cx is not in I and then g(x) would not be continiouse because it would not be defined at cx. is this proof blueprint correct?)

my real question has to do with the converse. I am confused what the converse would be and i am not sure if it is true. any help would be sweet.

Explanation / Answer

my real question has to do with the converse. I am confused what the converse would be and i am not sure if it is true. any help would be sweet.

You're correct about the proof. The converse would follow a similar reasoning:

Let I be a nonempty subset of R, let f:I->R be a function. Let c be a member of the real numbers and define the function g:I->R by g(x)=f(cx), for x in I. If g(x) is continuous, then f(x) is continuous.

How do we prove this?

Use the same case method as your original proof.

(The converse would also be true.)

Case 1: cx is in I; f(x) is continuous because f(x) is continuous at every point in I.

Case 2: cs is not in I; f(x) is continuous because it defined as f(cx) = g(x) and g(x) is continuous at every point in I.

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