Continuity (intermediate value theorem)

[mtayab1994]

Homework Statement

Let f be a continuous function on the interval I=[a,b] such that for every x in [a,b] f(x)≠0.

Show that the function f(x) doesn't change its sign.( like increasing or decreasing)

The Attempt at a Solution

Well for this to be true, we need to have f(a)>0 and f(b)>0 and f(x) is increasing so then it won't change the monotony. If we have f(a)<0 and f(b)<0, then f(x) is decreasing, hence we will not find any x in the interval I such that f(x)=0. Therefore for every x in the interval I f(x)≠0. Am I correct? Do I need to explain a bit more or what?

 

[jgens]

Well for this to be true, we need to have f(a)>0 and f(b)>0 and f(x) is increasing so then it won't change the monotony. If we have f(a)<0 and f(b)<0, then f(x) is decreasing, hence we will not find any x in the interval I such that f(x)=0. Therefore for every x in the interval I f(x)≠0. Am I correct? Do I need to explain a bit more or what?

This is nonsense. If $f$ does change sign, then there exist real numbers $c,d \in [a,b]$ such that $f(c) < 0$ and $0 < f(d)$. Now use the intermediate value theorem to derive a contradiction.

 

[HallsofIvy]

Homework Statement

Let f be a continuous function on the interval I=[a,b] such that for every x in [a,b] f(x)≠0.

Show that the function f(x) doesn't change its sign.( like increasing or decreasing)

Did you add the parenthetical statement? I believe you are misunderstanding the question! A function is "increasing or decreasing" if and only if the derivative of f doesn't change sign. $f(x)= x^2+ 1$, for example is continuous and never 0 on [-1, 1] but it certainly is NOT always increasing or always decreasing.

Use a proof by contradicction. If f(x1)> 0 and f(x2)< 0 (f changes sign) and f is continuous, what does the intermediate value theorem tell you?

The Attempt at a Solution

Well for this to be true, we need to have f(a)>0 and f(b)>0 and f(x) is increasing so then it won't change the monotony. If we have f(a)<0 and f(b)<0, then f(x) is decreasing, hence we will not find any x in the interval I such that f(x)=0. Therefore for every x in the interval I f(x)≠0. Am I correct? Do I need to explain a bit more or what?

 

[mtayab1994]

Did you add the parenthetical statement? I believe you are misunderstanding the question! A function is "increasing or decreasing" if and only if the derivative of f doesn't change sign. $f(x)= x^2+ 1$, for example is continuous and never 0 on [-1, 1] but it certainly is NOT always increasing or always decreasing.

Use a proof by contradicction. If f(x1)> 0 and f(x2)< 0 (f changes sign) and f is continuous, what does the intermediate value theorem tell you?

If f(x1)>0 and f(x2)<0 that entails f(x1)*f(x2)<0 so that means that there at least exists a number c such that f(x2)<f(c)=0<f(x1). Correct right?