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At the point $x=1$, the given function $f(x)=\left\{\begin{array}{l}
x^3-1 ; 1 \lt x \lt \infty \\
x-1 ;-\infty \lt x \leq 1
\end{array}\right.$ is
Options:
x^3-1 ; 1 \lt x \lt \infty \\
x-1 ;-\infty \lt x \leq 1
\end{array}\right.$ is
Solution:
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Verified Answer
The correct answer is:
Continuous and not differentiable
We have $R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{\left\{(1+h)^3-1\right\}-0}{h}=3$
$L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{\{(1-h)-1\}-0}{-h}=1$
$\therefore R f^{\prime}(1) \neq L f^{\prime}(1) \Rightarrow f(x)$ is not differentiable at $x=1$.
Now,$f(1+0)=\lim _{h \rightarrow 0} f(1+h)=0$
and $f(1-0)=\lim _{h \rightarrow 0} f(1-h)=0$
$\therefore f(1+0)=f(1-0)=f(0) \Rightarrow f(x)$ is continuous at $x=1$. Hence at $x=1, f(x)$ is continuous and not differentiable.
$=\lim _{h \rightarrow 0} \frac{\left\{(1+h)^3-1\right\}-0}{h}=3$
$L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{\{(1-h)-1\}-0}{-h}=1$
$\therefore R f^{\prime}(1) \neq L f^{\prime}(1) \Rightarrow f(x)$ is not differentiable at $x=1$.
Now,$f(1+0)=\lim _{h \rightarrow 0} f(1+h)=0$
and $f(1-0)=\lim _{h \rightarrow 0} f(1-h)=0$
$\therefore f(1+0)=f(1-0)=f(0) \Rightarrow f(x)$ is continuous at $x=1$. Hence at $x=1, f(x)$ is continuous and not differentiable.
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