Q. A - Calculus

Question

For each given function f, find a real number a that makes f continuous at $x=0,$if possible.

$$\begin{gathered} \left(a\right) f\left(x\right)=\left\{\begin{array}{ll}3x+1,& \mathrm{if} x<0\\ 2x+a,& \mathrm{if} x\ge 0\end{array}\right. \\ \left(b\right) f\left(x\right)=\left\{\begin{array}{ll}\frac{a}{x+2},& \mathrm{if} x<0\\ 3,& \mathrm{if} x=0\\ ax+1,& \mathrm{if} x>0\end{array}\right. \end{gathered}$$

Step-by-Step Solution

Answer

(a) The function $f\left(x\right)=\left\{\begin{array}{ll}3x+1,& \mathrm{if} x<0\\ 2x+a,& \mathrm{if} x\ge 0\end{array}\right.$is continuous when $a=1.$

(b) The function $f\left(x\right)=\left\{\begin{array}{ll}\frac{a}{x+2},& \mathrm{if} x<0\\ 3,& \mathrm{if} x=0\\ ax+1,& \mathrm{if} x>0\end{array}\right.$cannot continue at any value of a.

Part (a) Step 1. Given information.

The given function is $f\left(x\right)=\left\{\begin{array}{ll}3x+1,& \mathrm{if} x<0\\ 2x+a,& \mathrm{if} x\ge 0\end{array}.\right.$

Part (a) Step 2. Continuity of the function.

The function is continuous when the left-hand limit, right-hand limit, and $f\left(0\right)$are the same.

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=f\left(0\right) \\ \underset{x\to 0}{\lim }(3x+1)=2\left(0\right)+a \\ 3\left(0\right)+1=a \\ a=1 \end{gathered}$$

so the function is continuous when $a=1.$

Part (b) Step 1. Given information.

The given function is $f\left(x\right)=\left\{\begin{array}{ll}\frac{a}{x+2},& \mathrm{if} x<0\\ 3,& \mathrm{if} x=0\\ ax+1,& \mathrm{if} x>0\end{array}.\right.$

Part (b) Step 2. Continuity of the function.

The function is continuous when the left-hand limit, right-hand limit, and $f\left(0\right)$are the same.

Equate the left-hand limit with $f\left(0\right).$

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=f\left(0\right) \\ \underset{x\to 0^{-}}{\lim }\frac{a}{x+2}=3 \\ \frac{a}{0+2}=3 \\ a=6 \end{gathered}$$

Equate the right-hand limit with $f\left(0\right).$

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right) \\ \underset{x\to 0^{+}}{\lim }(ax+1)=3 \\ a\left(0\right)+1=3 \\ 1=3 \end{gathered}$$

the equation is absurd, so the function is not continuous at any value of a.