Question

Let
$f\left(x\right)=\{\begin{array}{cc}{x}^2& \text{if}x\le x_0\\ ax+b& \text{if}x>x_0\end{array}$
The values of the coefficients a and b for which the function is continuous and has a derivative at $x_0$. are

• A. $a=x_0$, $b=-x_0$
• B. $a=2x_0$, $b=-x_0^2$
• C. $a=x_0^2$, $b=-x_0$
• D. $a=x_0$, $b=-x_0^2$

Answer 1

The correct option is B $a=2x_0$, $b=-x_0^2$

For $f$ to be continuous everywhere, we must have
$x_0^2=f\left(x_0\right)=\underset{x\to x_{0^{+}}}{lim}f\left(x\right)=a{x}_0+b$
Also, $f$ has a derivative at $x_0$ if $f^{\prime }\left(x_{0^{+}}\right)=f^{\prime }\left(x_{0^{-}}\right)$,
Now, $f^{\prime }\left(x_{0^{+}}\right)=\underset{h\to 0^{+}}{lim}\frac{f(x_0+h)-f\left(x_0\right)}{h}$
$=\underset{h\to 0^{+}}{lim}\frac{a(x_0+h)+b-x_0^2}{h}$
$=\underset{h\to 0^{+}}{lim}(\frac{a{x}_0+b-x_0^2}{h}+a)=\underset{h\to 0^{+}}{lim}a=a$ $[\because x_0^2=a{x}_0+b]$
and $f^{\prime }\left(x_{0^{-}}\right)=\underset{h\to 0^{-}}{lim}\frac{f(x_0+h)-f\left(x_0\right)}{h}$
$=\underset{h\to 0^{-}}{lim}\frac{{(x_0+h)}^2-x_0^2}{h}$
$=\underset{h\to 0^{-}}{lim}\frac{x_0^2+h^2+2x_{0}h-x_0^2}{h}$
$=\underset{h\to 0^{-}}{lim}(h+2x_0)=2x_0$
Hence $a=2x_0$, and $b=x_0^2-a{x}_0=x_0^2-2x_0^2=-x_0^2$