TIFR2010-Maths-A-8

Question

Let $f(x)= |x|^{3/2}, x \in \mathbb{R}$. Then

• A. $f$ is uniformly continuous.
• B. $f$ is continuous, but not differentiable at $x=0$.
• C. $f$ is differentiable and $f ' $ is continuous.
• D. $f$ is differentiable, but $f ' $ is discontinuous at $x=0$.

Comments

Answer is option B because when we draw  graph of the given function there is a cusp at x=0 and the function is continuous at x=0. whenever we have cusp at a point then at that point the fuction is not differentiable..

Answer 1

More insight here,

http://www.wolframalpha.com/input/?i=y%3D|x|^(3%2F2)

I think answer should be B

Comments

Why left part is not imaginary like this

What will be the definition of this function

$$f(x) = \begin{cases} x^{3/2} &\text{ for }x \geq 0 \\ (-x)^{3/2} &\text{ for } x<0 \end{cases}$$

OR

$$f(x) = \begin{cases} x^{3/2} &\text{ for }x \geq 0 \\ -(x^{3/2}) &\text{ for } x<0 \end{cases}$$ ?

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i think the first breakdown of function is correct beacause we wont be able to take the square root of negative number...

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Any reference?

Answer 2

The answer is option ©

For first option, let us suppose that $f$ is uniformly continuous. Then, we have

$||x|^{3/2}-|y|^{3/2}|$ $\leq$ |x-y|

Since the domain is $\mathbb R$, let us choose our $x$ and $y$ as $4$ and $9$ then left-hand side of the above inequation will be $|19|$ and the right-hand side will be $|5|$. Clearly, the above inequation does not hold for $x=4$ and $y=9$, and in fact you can find many such $x$ and $y$. Hence $f$ is not uniformly continuous.

For the second option, $f$ is definitely continuous and for that, you can apply the simple definition (assuming you know that), going to differentiability, $|x|^{a}$ is differentiable if $a>1$ because you will always get a smooth graph at $x=0$ and hence it will be differentiable at $x=0$., therefore this option will be incorrect.

Going to third option, we will have to find the derivative function of $f(x)$ which will be

$$Df(x)=\begin{cases}3/2x^{½}, &\text{if $x\geq 0$}\\-3/2x^{1/2}, &\text{if $x< 0$}\end{cases}$$

You can easily verify with the help of definition of continuity that this function is continuous at $x=0$. Hence third option is correct and fourth option is incorrect.

Answer 3

f is differential & first derivative of f is discontinuous at x =0

Comments

can u solve it ..

Answer 4

The graph for $f(x) = |x|^{\frac{3}{2}}$ looks like :

And the plot of its derivative is :

Clearly : option D

f(x) is not continuous because the domain is all R, but the plot is only possible for positive real numbers.

Comments

Any differentiable function must be continuous at every point in its domain. 

Then how is option D possible?can you please explain?