Name: Example of Mean Value Theorem
Uploaded: 2019-07-19
Duration: 2 min 17 s
Description: You've already learned about the Mean Value Theorem. Let’s explore more about the theorem with some examples in this video. Watch and learn!

Question

Discuss continuity and differentiability of $y = f (x)$ in $[- 2, 5]$ , where [ .] denotes $GIF$ {.} denotes $FPF$

$f (x) = \{\begin{matrix} [x] & , & x \in [- 2, 0] \\ {x} & , & x \in (0, 2) \\ \frac{x^{2}}{4} & , & x \in [2, 3] \\ \frac{1}{\log_{4} (x - 3)} & , & x \in [3, 5] \end{matrix}$

Accepted Answer

Here,

$f (x) = \{\begin{matrix} [x] & , & x \in [- 2, 0] \\ {x} & , & x \in (0, 2) \\ \frac{x^{2}}{4} & , & x \in [2, 3] \\ \frac{1}{\log_{4} (x - 3)} & , & x \in [3, 5] \end{matrix}$

$f (x) = \{\begin{matrix} \begin{matrix} \begin{matrix} - 2 & , & x \in [- 2, - 1) \end{matrix} \\ \begin{matrix} - 1 & , & x \in [- 1, 0) \end{matrix} \\ \begin{matrix} 0 & , & x = 0 \end{matrix} \\ \begin{matrix} x & , & x \in (0, 1) \end{matrix} \\ \begin{matrix} 0 & , & x = 1 \end{matrix} \\ \begin{matrix} x - 1 & , & x \in (1, 2) \end{matrix} \\ \frac{x^{2}}{4}, x \in [2, 3) \\ \frac{1}{\log_{4} (x - 3)}, x \in [3, 5] \end{matrix} \end{matrix}$ $f^{'} (x) = \begin{matrix} \{\begin{matrix} \begin{matrix} 0 & , & x \in [- 2, - 1) \end{matrix} \\ \begin{matrix} 0 & , & x \in [- 1, 0) \end{matrix} \\ \begin{matrix} 0 & , & x = 0 \end{matrix} \\ \begin{matrix} 1 & , & x \in (0, 1) \end{matrix} \\ \begin{matrix} 0 & , & x = 1 \end{matrix} \\ \begin{matrix} 1 & , & x \in (1, 2) \end{matrix} \\ \begin{array}{l} \frac{x}{2}, x \in (2, 3) \\ \frac{1}{(x - 3) \ln (x - 3)} \begin{matrix} , & x \in (3, 5) \end{matrix} \end{array} \end{matrix} \begin{matrix} \end{matrix} \end{matrix}$ $\begin{matrix} \end{matrix}$

It can be seen that

$f (x)$ is discontinuous at $x = - 1, 0, 1, 3$

$f (x)$ is not defined at $x = 4,$ so it is discontinuous at $x = 4$

$f (x)$ is non differentiable at above points.

At $x = 2$

$LHD$ $\Rightarrow f^{'} (2^{-}) = 1$ and $RHD \Rightarrow f^{'} (2) = 1$

It is differentiable at $x = 2$

Hence, $f (x)$ is differentiable and continuous at $x = 1$ but neither differentiable nor continuous at $x = - 1, 0, 1, 3, 4$

Discuss continuity and differentiability of $y = f (x)$ in $[- 2, 5]$ , where [ .] denotes $GIF$ {.} denotes $FPF$

$f (x) = \{\begin{matrix} [x] & , & x \in [- 2, 0] \\ {x} & , & x \in (0, 2) \\ \frac{x^{2}}{4} & , & x \in [2, 3] \\ \frac{1}{\log_{4} (x - 3)} & , & x \in [3, 5] \end{matrix}$

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Discuss continuity and differentiability of y=f(x) in [-2,5], where [ .] denotes GIF {.} denotes FPF fx =[x],x∈[-2,0]{x},x∈(0,2)x24,x∈[2,3]1log4(x-3),x∈[3,5]

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Discuss continuity and differentiability of $y = f (x)$ in $[- 2, 5]$ , where [ .] denotes $GIF$ {.} denotes $FPF$

$f (x) = \{\begin{matrix} [x] & , & x \in [- 2, 0] \\ {x} & , & x \in (0, 2) \\ \frac{x^{2}}{4} & , & x \in [2, 3] \\ \frac{1}{\log_{4} (x - 3)} & , & x \in [3, 5] \end{matrix}$