Question

let $f(x)=\

$$\begin{cases}-2x+b & \\text{if } x<2 \\\\ \\frac{-24}{x-b} & \\text{if } x\\geq2\\end{cases}$$

$
there are exactly two values for $b$ which make $f(x)$ a continuous function at $x = 2$. the one with the greater absolute value is
$b=$
now for fun, try to graph $f(x)$.
submit question

Explanation:

Step1: Set left limit at $x=2$

For $x<2$, $f(x)=-2x+b$. Compute limit as $x\to2^-$:
$\lim_{x\to2^-} f(x) = -2(2) + b = b - 4$

Step2: Set right limit at $x=2$

For $x\geq2$, $f(x)=\frac{-24}{x-b}$. Compute limit as $x\to2^+$:
$\lim_{x\to2^+} f(x) = \frac{-24}{2 - b}$

Step3: Equate limits for continuity

For continuity, left = right limit:
$b - 4 = \frac{-24}{2 - b}$

Step4: Solve the equation

Multiply both sides by $(2 - b)$:
$(b - 4)(2 - b) = -24$
Expand left side:
$-b^2 + 6b - 8 = -24$
Rearrange to quadratic form:
$b^2 - 6b - 16 = 0$
Factor:
$(b - 8)(b + 2) = 0$
Solutions: $b=8$ or $b=-2$

Step5: Select value with greater absolute value

$|8|=8$, $|-2|=2$. $8>2$.

Answer:

$8$