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.

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{ax}-\mathrm{b},\mathrm{if}\mathrm{x}<2\\ {\mathrm{bx}}^2+1,\mathrm{if}\mathrm{x}\ge 2\end{array}\right.$

Short answer

$\mathrm{Ans}:\mathrm{b}=\frac{1}{5}\&\mathrm{a}=\frac{4}{5}$

Step-by-step solution

Step 1. Given information is:

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{ax}-\mathrm{b},\mathrm{if}\mathrm{x}<2\\ {\mathrm{bx}}^2+1,\mathrm{if}\mathrm{x}\ge 2\end{array}\right.$

Step 2. Calculating a and b

$$\begin{gathered} \mathrm{Differentiating}\mathrm{the}\mathrm{function}, \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{a},\mathrm{if}\mathrm{x}<2\\ 2\mathrm{bx},\mathrm{if}\mathrm{x}\ge 2\end{array}\right. \\ \mathrm{Since}\mathrm{it}\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{differentiable}\mathrm{therefore}\mathrm{left}\mathrm{hand}\mathrm{derivative}\mathrm{at}\mathrm{x}=2, \\ \mathrm{is}\mathrm{a}\mathrm{and}\mathrm{right}\mathrm{hand}\mathrm{derivative}\mathrm{at}\mathrm{x}=2\mathrm{is}\mathrm{f}\text{'}\left(2\right)=4\mathrm{b}. \\ \mathrm{Therefore}, \\ \mathrm{a}=4\mathrm{b}.....\left(1\right) \\ \mathrm{Now}\mathrm{the}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{at}\mathrm{x}=2 \\ \mathrm{Therefore}, \\ \underset{\mathrm{x}\to 2^{-}}{\lim }\mathrm{ax}+\mathrm{b}=\underset{\mathrm{x}\to 2^{+}}{\lim }{\mathrm{bx}}^2+1 \\ 2\mathrm{a}+\mathrm{b}=4\mathrm{b}+1 \\ 2\left(4\mathrm{b}\right)+\mathrm{b}=4\mathrm{b}+1[\mathrm{From}(1\left)\right] \\ 5\mathrm{b}=1 \\ \mathrm{b}=\frac{1}{5} \\ \mathrm{Put}\mathrm{value}\mathrm{of}\mathrm{b}\mathrm{in}\left(1\right) \\ \mathrm{a}=\frac{4}{5} \end{gathered}$$