Question

Let $f\left(x\right)=$ $\{\begin{array}{c}{x}^2,\text{if}x\le x_0\\ ax+b,\text{if}x>x_0\end{array}$. If $f$ is differentiable at $x_0$, then

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

Answer 1

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

Let $f\left(x\right)={\{}_{ax+b,ifx>x_0}^{x^2,ifx\le x_0}L{f}^{{}^{\prime }}\left(x_0\right)={lim}_{h⟶0}\frac{f(x_0-h)-f\left(x_0\right)}{-h}={lim}_{h⟶0}\frac{{(x_0-h)}^2-x_0^2}{-h}={lim}_{h⟶0}\frac{(x_0-h+x_0)(x_0-h-x_0)}{-h}={lim}_{h⟶0}\frac{-h(2x_0-h)}{-h}={lim}_{h⟶0}2x_0--h=2x_0$

$R{f}^{{}^{\prime }}\left(x_0\right)={lim}_{h⟶0}\frac{f(x_0+h)-f\left(x_0\right)}{-h}={lim}_{h⟶0}\frac{a(x_0+h)+b-x_0^2}{-h}={lim}_{h⟶0}\frac{a{x}_0+ah+b-x_0^2}{-h}$

Since $f$ is differentiable at$x_0$, $R{f}^{{}^{\prime }}\left(x_0\right)$ must be finite, which is only possible when

$a{x}_0+ah+b-x_0^2=0$ when $h\to 0$

i.e $a{x}_0+b-x_0^2=0\to \left(1\right)$

Now,

$R{f}^{{}^{\prime }}\left(x_0\right)={lim}_{h⟶0}\frac{a{x}_0+ah+b-x_0^2}{-h}$

Applying L'Hospital Rule,

$R{f}^{{}^{\prime }}\left(x_0\right)={lim}_{h⟶0}\frac{0+a+0-0}{1}R{f}^{{}^{\prime }}\left(x_0\right)==a$

Now, since $f$ is differentiable at $x_0$

$L{f}^{{}^{\prime }}\left(x_0\right)=R{f}^{{}^{\prime }}\left(x_0\right)2x_0=aor,a=2x_0$

Substituting $a=2x_0$ in equation $\left(1\right)$

$a{x}_0+b-x_0^2=02x_0^2+b-x_0^2=0b=-x_0^2$

Answer $a=2x_0,b=-x_0^2$