Q. 21 - Calculus

Question

If possible, find constants a and b so that the function f that follows is continuous and differentiable everywhere. If it is not possible, explain why not.

$\mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}3x+a, \mathrm{if} \mathrm{x}<1\\ x^{\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} + 1, \mathrm{if} \mathrm{x}\ge 1\end{array}\right.$

Step-by-Step Solution

Answer

Ans. b = 6 & a = -2

Step 1. Given information is:

$\mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}3x+a, \mathrm{if} \mathrm{x}<1\\ x^{\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} + 1, \mathrm{if} \mathrm{x}\ge 1\end{array}\right.$

Step 2. Calculating a and b

$$\begin{gathered} \mathrm{Differentiating} \mathrm{the} \mathrm{function}, \\ \mathrm{f}\text{'}\left(\mathrm{x}\right) = \left\{\begin{array}{l}3, \mathrm{if} \mathrm{x}<1\\ \frac{\mathrm{b}}{2}{\mathrm{x}}^{\frac{\mathrm{b}}{2}-1}, \mathrm{if} \mathrm{x}\ge 1\end{array}\right. \\ \mathrm{Since} \mathrm{it} \mathrm{is} \mathrm{continuous} \mathrm{and} \mathrm{differentiable} \mathrm{therefore} \mathrm{left} \mathrm{hand} \mathrm{derivative} \mathrm{at} \mathrm{x}=1, \\ \mathrm{is} 3 \mathrm{and} \mathrm{right} \mathrm{hand} \mathrm{derivative} \mathrm{at} \mathrm{x}=1 \mathrm{is} \mathrm{f}\text{'}\left(1\right)=\frac{\mathrm{b}}{2}{\left(1\right)}^{\frac{\mathrm{b}}{2}-1} \mathrm{Therefore}, \\ 3=\frac{\mathrm{b}}{2} \\ \mathrm{b}=6 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{function} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as}, \\ \mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}3\mathrm{x}+\mathrm{a}, \mathrm{if} \mathrm{x}<1\\ {\mathrm{x}}^3 + 1, \mathrm{if} \mathrm{x}\ge 1\end{array}\right. \\ \mathrm{Now} \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x} = 1 \\ \underset{\mathrm{x}\to 1^{-}}{\lim }3\mathrm{x}+\mathrm{a} = \underset{\mathrm{x}\to 1^{+}}{\lim }{\mathrm{x}}^3 \\ 3+\mathrm{a}=1 \\ \mathrm{a}=-2 \end{gathered}$$