Question

let
f(x)=\begin{cases}b - 2x&\text{if }x<2\\-\frac{24}{x - b}&\text{if }xgeq2end{cases}
find the two values of (b) for which (f) is a continuous function at (2).
the one with the greater absolute value is (b = 3+sqrt{41}).
now draw a graph of (f).

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = 2$, $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)$. Calculate the left - hand limit $\lim_{x
ightarrow2^{-}}f(x)$ and the right - hand limit $\lim_{x
ightarrow2^{+}}f(x)$.
$\lim_{x
ightarrow2^{-}}f(x)=b - 2\times2=b - 4$.
$\lim_{x
ightarrow2^{+}}f(x)=-\frac{24}{2 - b}$.

Step2: Set the two limits equal

Set $b - 4=-\frac{24}{2 - b}$. Cross - multiply to get $(b - 4)(2 - b)=- 24$.
Expand the left - hand side: $2b-b^{2}-8 + 4b=-24$.
Rearrange to get a quadratic equation: $b^{2}-6b - 16 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-6$, $c=-16$), use the quadratic formula $b=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
$b=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times(-16)}}{2\times1}=\frac{6\pm\sqrt{36 + 64}}{2}=\frac{6\pm\sqrt{100}}{2}=\frac{6\pm10}{2}$.
The two solutions are $b=\frac{6 + 10}{2}=8$ and $b=\frac{6-10}{2}=-2$.
The one with the greater absolute value is $b = 8$.

Answer:

$8$