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

  1. $f$ is uniformly continuous.
  2. $f$ is continuous, but not differentiable at $x=0$.
  3. $f$ is differentiable and $f ' $ is continuous.
  4. $f$ is differentiable, but $f ' $ is discontinuous at $x=0$.
asked in Calculus by Veteran (29k points)   | 187 views
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..

3 Answers

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f is differential & first derivative of f is discontinuous at x =0
answered by Veteran (45.3k points)  
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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.

answered by Loyal (4.1k points)  
0 votes

More insight here,

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

I think answer should be B

answered by (11 points)  

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

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

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