Question

find a and b such that f is differentiable everywhere.$f(x)=\begin{cases}5cos(x), & x < 0 \ax + b, & x geq 0end{cases}LXB0b = square$submit answer

Explanation:

Step1: Ensure continuity at $x=0$

For continuity, $\lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0)$
$\lim_{x\to0^-}5\cos(x) = 5\cos(0) = 5$
$\lim_{x\to0^+}(ax+b) = a\cdot0 + b = b$
Set equal: $b = 5$

Step2: Ensure differentiability at $x=0$

For differentiability, left and right derivatives at $x=0$ must be equal.
Left derivative: $\frac{d}{dx}[5\cos(x)]\bigg|_{x=0} = -5\sin(0) = 0$
Right derivative: $\frac{d}{dx}[ax+b]\bigg|_{x=0} = a$
Set equal: $a = 0$

Answer:

$a = 0$
$b = 5$