Continuity and Differentiation...

• Jun 21st 2008, 08:58 AM
Vedicmaths
Continuity and Differentiation...
Hello,

I need help in figuring out the continuity of these weird looking function..

Q1) Find the maximum and minimum value of the function:
f(x) = 4x^3 - x^2 - 4x + 2 on [ -1,1] and on [0,1]?

Q2) Let f: [0, infinity), -> R be defined by

f(x) = x sin (1/x) when x>0
and f(x) = 0 when x = 0.
Figure out if 'f' is continuous and differentiable at x = 0.

Q3) Define f: [-2,2] -> R by f(x) = |x^3 - 1|.

f: [-2,2] -> R by f(x) = |x^3| -1.
Determine the points where f is differentiable and find the derivative at those points?
• Jun 21st 2008, 09:39 AM
galactus
Quote:

Q2) Let f: [0, infinity), -> R be defined by

f(x) = x sin (1/x) when x>0
and f(x) = 0 when x = 0.
Figure out if 'f' is continuous and differentiable at x = 0.
Well, to be continuous the limit must exist.

$\displaystyle \lim_{x\to 0^{+}}xsin(\frac{1}{x})$

$\displaystyle -1 \leq sin(1/x) \leq 1$

so $\displaystyle -x \leq xsin(1/x) \leq x$

But $\displaystyle -x \;\ and \;\ x\rightarrow{0} \;\ as \;\ x\rightarrow{0^{+}}$

This, $\displaystyle \lim_{x\to 0^{+}}xsin(1/x) = 0$

Let's describe the interval where it is continuous.

If we have:

$\displaystyle f(x) = \begin{Bmatrix}xsin(1/x), \;\ x\neq 0\\0, \;\ x=0\end{Bmatrix}$

By the Squeezing Theorem:

$\displaystyle -|x|\leq xsin(1/x) \leq |x|$

And we can conclude that $\displaystyle \lim_{x\to 0}f(x)=0$

So f is continuous on the entire real line.
• Jun 21st 2008, 09:53 AM
Mathstud28
Quote:

Originally Posted by Vedicmaths
Hello,

I need help in figuring out the continuity of these weird looking function..

Q1) Find the maximum and minimum value of the function:
f(x) = 4x^3 - x^2 - 4x + 2 on [ -1,1] and on [0,1]?

Q2) Let f: [0, infinity), -> R be defined by

f(x) = x sin (1/x) when x>0
and f(x) = 0 when x = 0.
Figure out if 'f' is continuous and differentiable at x = 0.

Q3) Define f: [-2,2] -> R by f(x) = |x^3 - 1|.

f: [-2,2] -> R by f(x) = |x^3| -1.
Determine the points where f is differentiable and find the derivative at those points?

For the limit, it also might be more apparent to let

$\displaystyle \varphi=\frac{1}{x}$

So as $\displaystyle x\to{0}\Rightarrow\varphi\to\infty$

So we have

$\displaystyle \lim_{\varphi\to\infty}\frac{\sin(\varphi)}{\varph i}$

Now it might be a little more obvious
• Jun 21st 2008, 11:01 AM
TheEmptySet
Quote:

Originally Posted by Vedicmaths
Hello,

Q2) Let f: [0, infinity), -> R be defined by

f(x) = x sin (1/x) when x>0
and f(x) = 0 when x = 0.
Figure out if 'f' is continuous and differentiable at x = 0.

We can use the limit defintion to see if it is differentiable at x=0

$\displaystyle \lim_{x \to 0}\frac{f(x)-f(0)}{x-0}$

$\displaystyle \lim_{x \to 0}\frac{x\sin\left( \frac{1}{x}\right)-0}{x-0}=\lim_{x \to 0}\sin\left( \frac{1}{x}\right)$

This limit does dont exits so the function is not differentiable at x=0
• Jun 21st 2008, 02:15 PM
Vedicmaths
Thanks everyone for the solution.

Regards,
• Jun 21st 2008, 02:28 PM
galactus
I am sorry, but I beg to differ about the sine limit. It appears to be continuous at 0. The oscillations are damped by the factor x.

The limit is 0 rather approaching from the left or right.

sin(1/x) is not continuous, though. You can see that from the oscillation on the graph.
• Jun 21st 2008, 02:29 PM
galactus
Here is the graph of sin(1/x). It is not continuous at x=0
• Jun 21st 2008, 03:03 PM
Plato
Quote:

Originally Posted by Vedicmaths
Thanks everyone for the solution.