Answer :
Therefore the f is continuous on R , i.e. real numbers and f is not differentiable at 0 ,Option A and D is the correct answer.
What is a Function ?
A function is a mathematical statement which find relation between independent variable and dependent variable.
It always comes with a defined range and domain.
It is given that :
[tex]\rm f(x) = x^{1/3}\\\\LHD (at\; x = 0) = RHD (at \; x = 0)\\\\\= \lim_{x\rightarrow 0}\dfrac{f(x)-f(0)}{x-0}\\\\=\lim_{h\rightarrow 0}\dfrac{f(0-h)-f(0)}{0-h-0}[/tex]
[tex]=\lim_{h\rightarrow 0}\dfrac{f(0-h)-f(0)}{0-h-0} \\\\= \lim_{h\rightarrow 0}\dfrac{(-h)^{1/3}}{-h}\\\\\lim_{h\rightarrow 0}\dfrac{(-1)^{1/3}(h)^{1/3}}{-1*h}\\\\\lim_{h\rightarrow 0} (-1)^{-2/3}h^{-2/3}[/tex]
if we substitute h = 0 ,
Here, LHD and RHD does not exist at x = 0 So, f(x) is not differentiable at x = 0 .
Therefore the f is continuous on R , i.e. real numbers
and f is not differentiable at 0
Option A and D is the correct answer.
To know more about Function
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