Question

If the function $f\left(x\right)=\{\begin{array}{cc}ax+b& x\le -1\\ a{x}^3+x+2b& x>-1\end{array}$ is differentiable for all values of x then

• A. $a=\frac{1}{2}$
• B. $b=0$
• C. $a=-\frac{1}{2}$
• D. $b=1$

Answer 1

Answer: (C), (D)

$b=1$
$a=-\frac{1}{2}$

$f\left(x\right)=\{\begin{array}{cc}ax+b& x\le -1\\ a{x}^3+x+2b& x>-1\end{array}$

$-a+b=-a-1+2b$ (As $f\left(x\right)$ is continues at $x=-1$)

So $b=1$

$f^{{}^{\prime }}\left(x\right)=\{\begin{array}{cc}a& x\le -1\\ 3a{x}^2+1& x>-1\end{array}$

So $a=3a+1$ as $f\left(x\right)$ is differentiable at $x=-1$

$a=\frac{-1}{2}$