When a **function** is **differentiable** it is also **continuous**. But a **function** can be **continuous** but **not differentiable**

**f(x)=x+1 is both continuous and differential**

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what value you are getting for the function when $x=1 ?$

If it is $x \neq 1 $ and $f(x)$ is defined for all reals i.e. $\mathbb{R} \backslash \{1\}$ then $f(x)$ must be continuous or if you write $f(x) = \frac{x^2 – 1}{x-1}, x \neq 1$ and $f(x) = 2, x=1$ then you can say $f(x)$ is continuous for all real values of $x$ or when your function is defined as $f(x) = x+1$ for all $x \in \mathbb{R}$ then also you can say $f(x)$ is continuous.

But here in the question domain of the function is not defined. if the domain is $x \in \mathbb{R}$ If we have to check the continuity for $x =1$ then the given function is not continuous because $f(1)$ is undefined.

If it is $x \neq 1 $ and $f(x)$ is defined for all reals i.e. $\mathbb{R} \backslash \{1\}$ then $f(x)$ must be continuous or if you write $f(x) = \frac{x^2 – 1}{x-1}, x \neq 1$ and $f(x) = 2, x=1$ then you can say $f(x)$ is continuous for all real values of $x$ or when your function is defined as $f(x) = x+1$ for all $x \in \mathbb{R}$ then also you can say $f(x)$ is continuous.

But here in the question domain of the function is not defined. if the domain is $x \in \mathbb{R}$ If we have to check the continuity for $x =1$ then the given function is not continuous because $f(1)$ is undefined.

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