# If f has the IVP, prove kf does too

• k3k3

## Homework Statement

Suppose f has the intermediate value property on an interval I and let k be a constant. Prove that kf has the intermediate value property.

## The Attempt at a Solution

Since f has IVP on I, then there is distinct a and b in I and f(c)=v exists and is between f(a) and f(b).

Let k be a constant real number.

Then kf(a), kf(b) and kf(c) are all real numbers where kf(c) is between kf(a) and kf(b) by properties of real numbers.

So then kf has IVP on I?

k3k3 said:

## Homework Statement

Suppose f has the intermediate value property on an interval I and let k be a constant. Prove that kf has the intermediate value property.

## The Attempt at a Solution

Since f has IVP on I, then there is distinct a and b in I and f(c)=v exists and is between f(a) and f(b).
No, that is not the intermediate vaue theorem. Rather, for any a and b, if v is any number between f(a) and f(b), then there exists c, between a and b, such that f(c)= b.
Let k be a constant real number.

Then kf(a), kf(b) and kf(c) are all real numbers where kf(c) is between kf(a) and kf(b) by properties of real numbers.

So then kf has IVP on I?
You are "going the wrong way", assuming from the beginning that your number is of the form kf(x). That is, basically, what you want to prove.

Let a and b be any two numbers and v be a number between kf(a) and kf(b). Then, if k> 0, v/k lies between f(a) and f(b) so, because f has the intermediate value property, there exist c such that f(c)= v/k. Then kf(c)= k(v/k)= v and we are done.

You do the cases (a) k< 0 and (b) k= 0.