1. Tardigrade
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  3. Mathematics
  4. At x=1, the function f(x)= begincasesx3-1 ; 1< x< ∞ x-1 ;-∞< x ≤ 1 text is endcases

Q. At $x=1$, the function $f(x)=\begin{cases}x^{3}-1 ; 1< x< \infty \\ x-1 ;-\infty< x \leq 1 \text { is }\end{cases}$

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Solution:

At $\left.x=1,\left(x^{3}-1\right)\right]_{x=1}=0$ and $\left.(x-1)\right]_{x=1}=0 $
$\therefore $ values of both branches match at $x=1 $
$ \therefore f(x)$ is continuous at $x=1$
$\left.\left.\left(x^{3}-1\right)^{\prime}\right]_{x=1}=3 x^{2}\right]_{x=1}=3$
and $\left.(x-1)^{\prime}\right]_{x=1}=1 $
$ \therefore $ Slopes of both branches don't match at $x=1 $
$ \therefore f(x)$ is not differentiable

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