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  • The function f(x) is defined as <img alt="f(x)=\left\{\begin{array}{cc} \frac{1}{3}-x & , \quad x<\frac{1}{3} \\ \left(\frac{1}{3}-x\right)^2, & x \geq \frac{1}{3} \end{array}\rig

The function f(x) is defined as
f(x)=\left\{\begin{array}{cc} \frac{1}{3}-x & , \quad x<\frac{1}{3} \\ \left(\frac{1}{3}-x\right)^2, & x \geq \frac{1}{3} \end{array}\right.
then in the interval (0,1), the Mean Value Theorem is not true because
 

Option: 1

 f(x) is not continuous
 


Option: 2

 f(x) is not differentiable
 


Option: 3

 f(0) \neq f(1)
 


Option: 4

none


Answers (1)

best_answer

 The function given is not differentiable at \mathrm{x=\frac{1}{3} } in [0,1] as \mathrm{L f^{\prime}\left(\frac{1}{3}\right)=-1 }and R \mathrm{f^{\prime}\left(\frac{1}{3}\right)=0. }
 Hence it does not satisfy the mean value theorem.

Posted by

Ajit Kumar Dubey

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