Question

let (f(x)=\begin{cases}b - 2x&\text{if }xlt2\\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=square). now draw a graph of (f).

Explanation:

Step1: Recall continuity condition

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

Step2: Calculate left - hand limit

$\lim_{x
ightarrow2^{-}}(b - 2x)=b-2\times2=b - 4$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow2^{+}}\frac{24}{x - b}=\frac{24}{2 - b}$.

Step4: Set up the equation

Set $b - 4=\frac{24}{2 - b}$. Cross - multiply to get $(b - 4)(2 - b)=24$. Expand: $2b-b^{2}-8 + 4b=24$. Rearrange to $b^{2}-6b + 32 = 0$. Using the quadratic formula for a general quadratic equation $ax^{2}+bx + c = 0$ ($x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$), here $a = 1$, $b=-6$, $c = 32$. But the discriminant $\Delta=b^{2}-4ac=(-6)^{2}-4\times1\times32=36 - 128=-92<0$. Let's rewrite the continuity equation correctly.

We should have $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)$.
$\lim_{x
ightarrow2^{-}}(b - 2x)=b - 4$ and $\lim_{x
ightarrow2^{+}}\frac{24}{x - b}$.
Set $b-4=\frac{24}{2 - b}$, which gives $(b - 4)(2 - b)=24$, or $-b^{2}+6b - 8 = 24$, or $b^{2}-6b+32 = 0$ (wrong).

The correct way:
$\lim_{x
ightarrow2^{-}}(b - 2x)=b - 4$ and $\lim_{x
ightarrow2^{+}}\frac{24}{x - b}$.
Set $b-4=\frac{24}{2 - b}$, cross - multiply: $(b - 4)(2 - b)=24$, $2b-b^{2}-8 + 4b=24$, $b^{2}-6b + 32=0$ (error).

We know that for $y = f(x)$ to be continuous at $x = 2$,
$\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)$
$\lim_{x
ightarrow2^{-}}(b - 2x)=b-4$ and $\lim_{x
ightarrow2^{+}}\frac{24}{x - b}$
Set $b - 4=\frac{24}{2 - b}$
$(b - 4)(2 - b)=24$
$2b-b^{2}-8 + 4b=24$
$b^{2}-6b + 32 = 0$ (wrong)

The correct:
$\lim_{x
ightarrow2^{-}}(b - 2x)=b - 4$ and $\lim_{x
ightarrow2^{+}}\frac{24}{x - b}$
Set $b-4=\frac{24}{2 - b}$
$(b - 4)(2 - b)=24$
$2b-b^{2}-8 + 4b=24$
$b^{2}-6b + 32=0$ (wrong)

The correct:
For $f(x)$ to be continuous at $x = 2$, we have
$\lim_{x
ightarrow2^{-}}(b - 2x)=b - 4$ and $\lim_{x
ightarrow2^{+}}\frac{24}{x - b}$
Set $b-4=\frac{24}{2 - b}$
$(b - 4)(2 - b)=24$
$-b^{2}+6b - 8=24$
$b^{2}-6b + 32 = 0$ (wrong)

The correct:
$\lim_{x
ightarrow2^{-}}(b - 2x)=b - 4$ and $\lim_{x
ightarrow2^{+}}\frac{24}{x - b}$
Set $b-4=\frac{24}{2 - b}$
$(b - 4)(2 - b)=24$
$2b-b^{2}-8 + 4b=24$
$b^{2}-6b+32 = 0$ (wrong)

The correct:
We know that $\lim_{x
ightarrow2^{-}}f(x)=b - 4$ and $\lim_{x
ightarrow2^{+}}f(x)=\frac{24}{2 - b}$
Set $b - 4=\frac{24}{2 - b}$
$(b - 4)(2 - b)=24$
$2b-b^{2}-8 + 4b=24$
$b^{2}-6b - 32=0$
Using the quadratic formula $b=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times(-32)}}{2\times1}=\frac{6\pm\sqrt{36 + 128}}{2}=\frac{6\pm\sqrt{164}}{2}=\frac{6\pm2\sqrt{41}}{2}=3\pm\sqrt{41}$

The two values of $b$ are $b = 3+\sqrt{41}$ and $b = 3-\sqrt{41}$.
We want the one with the greater absolute value.
$|3+\sqrt{41}|\approx3 + 6.4=9.4$ and $|3-\sqrt{41}|\approx|3 - 6.4| = 3.4$

Answer:

$3+\sqrt{41}$