Let $f(x)$ be a real valued function. If $f^{\prime}(x)$ is a constant for all $x \in \mathbb{R}, f(0)=2$…
Let $f(x)$ be a real valued function. If $f^{\prime}(x)$ is a constant for all $x \in \mathbb{R}, f(0)=2$ and $f^{\prime}(0)=1$, then
$\mathrm{f}(\mathrm{x})$ is not continuous on $\mathbb{R}$
$f(x)$ is continuous at $x=0,1,2$ and 3 only
$f(x)$ is continuous only on $[0, \infty)$
$\mathrm{f}(\mathrm{x})$ is continuous on $\mathbb{R}$
Solution
Since $f^{\prime}(x)$ is a constant
$
\therefore f^{\prime}(x)=a \text { (say) }...(1)
$
$\Rightarrow f(x)=a x+b$ where $b$ is arbitrary constant. ...(2)
Since $f(0)=2 \Rightarrow b=2$
Since $f^{\prime}(0)=1 \Rightarrow a=1$
$
\therefore f(x)=(x+2)
$
Which is continuous on $(-\infty, \infty)$ i.e $\mathbb{R}$