Question

find the values of $m$ and $b$ that make $f$ continuous everywhere:

$$f(x)=\
LATEXBLOCK0
$$

then showing your work, be sure to include all limit calculations and a sketch of the graph.

Explanation:

Step1: Simplify the first piece

First, factor the numerator of the first piece:
$32+4x-x^2 = -(x^2-4x-32) = -(x-8)(x+4)$
Then simplify the fraction:
$\frac{32+4x-x^2}{x+4} = \frac{-(x-8)(x+4)}{x+4} = -(x-8) = 8-x$ for $x
eq -4$

Step2: Ensure continuity at $x=-4$

For continuity, $\lim_{x \to -4^-} f(x) = f(-4)$
$\lim_{x \to -4^-} (8-x) = 8 - (-4) = 12$
$f(-4) = m(-4) + b = -4m + b$
Set equal: $-4m + b = 12$

Step3: Ensure continuity at $x=3$

First, calculate $\lim_{x \to 3^+} f(x)$:
$\lim_{x \to 3^+} 4 \cdot 2^{3-x} + 50 = 4 \cdot 2^{3-3} + 50 = 4 \cdot 1 + 50 = 54$
$f(3) = m(3) + b = 3m + b$
Set equal: $3m + b = 54$

Step4: Solve the system of equations

Subtract the first equation from the second:
$(3m + b) - (-4m + b) = 54 - 12$
$7m = 42$
$m = 6$
Substitute $m=6$ into $-4m + b = 12$:
$-4(6) + b = 12$
$-24 + b = 12$
$b = 36$

Answer:

$m=6$, $b=36$

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Sketch Notes (for reference):

1. For $x < -4$: Graph the line $y=8-x$ (a line with slope -1, y-intercept 8, with an open circle at $x=-4$, $y=12$)
2. For $x \in [-4,3]$: Graph the line $y=6x+36$ (slope 6, y-intercept 36), closed at both endpoints: $(-4,12)$ and $(3,54)$
3. For $x > 3$: Graph the exponential function $y=4 \cdot 2^{3-x} + 50$, which simplifies to $y=4 \cdot (\frac{1}{2})^{x-3} + 50$. It has a horizontal asymptote at $y=50$ as $x \to \infty$, and starts at an open circle at $(3,54)$ decreasing towards the asymptote.