# showing a function is continuous and differentiable

• Apr 22nd 2010, 05:16 PM
dgmath
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.
• Apr 22nd 2010, 05:21 PM
Drexel28
Quote:

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?
• Apr 22nd 2010, 05:35 PM
dgmath
Quote:

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?
• Apr 22nd 2010, 05:36 PM
Drexel28
Quote:

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\$
• Apr 22nd 2010, 05:40 PM
dgmath
Quote:

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?
• Apr 22nd 2010, 05:41 PM
Drexel28
Quote:

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)\$
• Apr 22nd 2010, 05:53 PM
dgmath
Quote:

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.
• Apr 22nd 2010, 06:00 PM
Drexel28
Quote:

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.