# Thread: showing a function is continuous and differentiable

1. ## showing a function is continuous and differentiable

Hello,

I have a problem in hand. I'll pose the question and include my attempt. Please correct me wherever I am wrong.

Let f and g are functions that satisfy |f(x)| =< |g(x)| for all x and such that g is not the zero function. Give examples of what g could be to make sure that 1) f is continuous at 0 2) f is differentiable at 0.

1) I couldn't solve it. any help?
2) If I chose function |f(x)| =< |g(x)| = x^2, for f to be differentiable at 0,
f'(0) = lim x->0 [f(x) - f(0)]/[x-0] = lim x->0 f(x)/x = limx->0 x^2/x = 0.
But on 2, am I even allowed to assume f(0) = 0?

I can rephrase things if there's a confusion. Thanks.

2. Originally Posted by dgmath
Hello,

I have a problem in hand. I'll pose the question and include my attempt. Please correct me wherever I am wrong.

Let f and g are functions that satisfy |f(x)| =< |g(x)| for all x and such that g is not the zero function. Give examples of what g could be to make sure that 1) f is continuous at 0 2) f is differentiable at 0.

1) I couldn't solve it. any help?
2) If I chose function |f(x)| =< |g(x)| = x^2, for f to be differentiable at 0,
f'(0) = lim x->0 [f(x) - f(0)]/[x-0] = lim x->0 f(x)/x = limx->0 x^2/x = 0.
But on 2, am I even allowed to assume f(0) = 0?

I can rephrase things if there's a confusion. Thanks.
1) $\displaystyle g(x)=x$
2) You know that $\displaystyle f(0)=0$. Why?

3. Originally Posted by Drexel28
1) $\displaystyle g(x)=x$
2) You know that $\displaystyle f(0)=0$. Why?
Is it because |g(x)| is non-zero, and since |f(x)|=<|g(x)|, it is valid to say that |f(0)| can be 0?

4. Originally Posted by dgmath
Is it because |g(x)| is non-zero, and since |f(x)|=<|g(x)|, it is valid to say that |f(0)| can be 0?
In your specific examples. If $\displaystyle |f(x)|\leqslant g(x)$ and if $\displaystyle g(0)=0$ then $\displaystyle |f(0)|\leqslant g(0)=0\implies |f(0)|=0\implies f(0)=0$

5. Originally Posted by Drexel28
In your specific examples. If $\displaystyle |f(x)|\leqslant g(x)$ and if $\displaystyle g(0)=0$ then $\displaystyle |f(0)|\leqslant g(0)=0\implies |f(0)|=0\implies f(0)=0$
That is logical. However, what has confused me in this question is the part when it says "g is not the zero function". I kind of thought value of g(x) can never be 0? Did I completely misunderstand it?

6. Originally Posted by dgmath
That is logical. However, what has confused me in this question is the part when it says "g is not the zero function". I kind of thought value of g(x) can never be 0? Did I completely misunderstand it?
$\displaystyle g(x)\ne 0(x)$

7. Originally Posted by Drexel28
$\displaystyle g(x)\ne 0(x)$
Great, thanks!

Just before I move on, for part 1), how do I say that for g(x) = x, f is continuous at 0? Because I figured my answer i.e. x^2 works too, but I dont know how its shown.

8. Originally Posted by dgmath
Great, thanks!

Just before I move on, for part 1), how do I say that for g(x) = x, f is continuous at 0? Because I figured my answer i.e. x^2 works too, but I dont know how its shown.
$\displaystyle f(0)=0$ by the same idea and the limit as x approaches zero of the function is zero by the squeeze theorem.