Question

for what values of (a) and (b) is the given function (f(x)) differentiable at (x = 0)?(f(x)=\begin{cases}ax + b&\text{if }x<0\\x^{2}+x&\text{if }xgeq0end{cases})(a=)(b =)

Explanation:

Step1: Check for continuity at $x = 0$

For a function to be differentiable at a point, it must be continuous at that point.
Left - hand limit as $x\to0^{-}$: $\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}(Ax + B)=B$.
Right - hand limit as $x\to0^{+}$: $\lim_{x\to0^{+}}f(x)=\lim_{x\to0^{+}}(x^{2}+x)=0$.
For continuity, $B = 0$.

Step2: Check for equal left - hand and right - hand derivatives at $x = 0$

The derivative of $y = Ax + B$ for $x<0$ is $y^\prime=A$.
The derivative of $y=x^{2}+x$ for $x\geq0$ is $y^\prime = 2x + 1$. The right - hand derivative at $x = 0$ is $2(0)+1=1$.
For differentiability, $A = 1$.

Answer:

$A = 1$
$B = 0$