Question

assignment 4: problem 10 (1 point) find a and b such that the function $f(x)=\begin{cases}x^{2}-4x + 2&\text{if }xleq - 2\ax + b&\text{if }x> - 2end{cases}$ is differentiable everywhere. $a = $ $b = $

Explanation:

Step1: Check continuity at $x = - 2$

For the function to be continuous at $x=-2$, $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{+}}f(x)$.
$\lim_{x
ightarrow - 2^{-}}(x^{2}-4x + 2)=(-2)^{2}-4\times(-2)+2=4 + 8+2 = 14$.
$\lim_{x
ightarrow - 2^{+}}(ax + b)=-2a + b$. So, $-2a + b=14$.

Step2: Check differentiability at $x=-2$

The derivative of $y=x^{2}-4x + 2$ is $y'=2x-4$. The derivative at $x = - 2$ is $2\times(-2)-4=-8$.
The derivative of $y=ax + b$ is $y'=a$. For the function to be differentiable at $x=-2$, $a=-8$.

Step3: Find the value of $b$

Substitute $a = - 8$ into $-2a + b=14$. We get $-2\times(-8)+b=14$, which simplifies to $16 + b=14$, so $b=-2$.

Answer:

$a=-8$
$b=-2$