Q. 8 - Calculus

Question

Given the following function $f$ , define $f\left(1\right)$ so that $f$ is continuous at $x=1$, if possible:

$f\left(x\right)=\left\{\begin{array}{ll}3x-1, if& x<1\\ x^2+1. if& x>1\end{array}\right.$

Step-by-Step Solution

Answer

As, $f\left(1^{-}\right)=f\left(1\right)=f\left(1^{+}\right)$. Therefore, $f$ is continuous at $x=1$.

Step 1. Given information

Function $f\left(x\right)=\left\{\begin{array}{ll}3x-1, if& x<1\\ x^2+1, if& x>1\end{array}\right.$

Step 2. Calculating f ( 1 - )

Function is continuous if $f\left(1^{-}\right)=f\left(1\right)=f\left(1^{+}\right)$.

$$\begin{gathered} f\left(1^{-}\right)=\underset{h\to 0}{\lim }\left(3\right(1-h)-1) \\ =\underset{h\to 0}{\lim }(3-3h-1) \\ =3-\underset{h\to 0}{3\lim }h-1 \\ =3-0-1 \\ =2 \\ So, f\left(1^{-}\right)=2 \end{gathered}$$

Step 3. Calculating f ( 1 + )

$$\begin{gathered} f\left(1^{+}\right)=\underset{h\to 0}{\lim }({(1+h)}^2+1)=\underset{h\to 0}{\lim }(1+h^2+2h+1) \\ =1+\underset{h\to 0}{\lim }{h}^2+2\underset{h\to 0}{\lim }h+1 \\ =1+0+0+1 \\ =2 \\ So, f\left(1^{+}\right)=2 \end{gathered}$$

Therefore, the value of $f\left(1\right)=2$ . As $f\left(1^{-}\right)=f\left(1\right)=f\left(1^{+}\right)$.