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Consider the following statements :
1- $f^{\prime}(2+0)$ does not exist.
2- $f^{\prime}(2-0)$ does not exist.
Let $f(x)$ be a function defined in $1 \leq x < \infty$ by $\quad[2014-I]$
$f(x)=\left\{\begin{array}{cc}2-x & \text { for } 1 \leq x \leq 2 \\ 3 x-x^{2} & \text { for } x>2\end{array}\right.$
Which of the above statements is/are correct?
Options:
1- $f^{\prime}(2+0)$ does not exist.
2- $f^{\prime}(2-0)$ does not exist.
Let $f(x)$ be a function defined in $1 \leq x < \infty$ by $\quad[2014-I]$
$f(x)=\left\{\begin{array}{cc}2-x & \text { for } 1 \leq x \leq 2 \\ 3 x-x^{2} & \text { for } x>2\end{array}\right.$
Which of the above statements is/are correct?
Solution:
2378 Upvotes
Verified Answer
The correct answer is:
1 only
$\mathrm{f}^{\prime}(2+0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}^{\prime}(2+\mathrm{h})$
$=\lim _{\mathrm{h} \rightarrow 0} 3-4-2 \mathrm{~h}=-1$
$\mathrm{f}^{\prime}(2-0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}^{\prime}(2-\mathrm{h})=(3-4-2 \mathrm{~h})=-1$
So, $\mathrm{f}^{\prime}(\mathrm{x})$ exist at $\mathrm{x}=2$
$=\lim _{\mathrm{h} \rightarrow 0} 3-4-2 \mathrm{~h}=-1$
$\mathrm{f}^{\prime}(2-0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}^{\prime}(2-\mathrm{h})=(3-4-2 \mathrm{~h})=-1$
So, $\mathrm{f}^{\prime}(\mathrm{x})$ exist at $\mathrm{x}=2$
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