Question

find the values of $m$ and $b$ that make $f$ continuous everywhere: $f(x)=\begin{cases}\frac{20 + x - x^{2}}{x + 4}&\text{if }x < - 4\mx + b&\text{if }xin-4,1\\5cdot2^{1 - x}+34&\text{if }x > 1end{cases}$. when showing your work, be sure to include all limit calculations and a sketch of the graph. $m = $ $b = $ question help: message instructor

Explanation:

Step1: Simplify the first - piece function

For \(x < - 4\), simplify \(\frac{20 + x-x^{2}}{x + 4}\). Factor the numerator: \(20 + x-x^{2}=-(x^{2}-x - 20)=-(x - 5)(x+4)\). Then \(\frac{20 + x - x^{2}}{x + 4}=\frac{-(x - 5)(x + 4)}{x + 4}=-x + 5\) for \(x
eq - 4\). Calculate \(\lim_{x
ightarrow - 4^{-}}f(x)=\lim_{x
ightarrow - 4^{-}}(-x + 5)=9\).

Step2: Use continuity at \(x=-4\)

For the function to be continuous at \(x=-4\), \(\lim_{x
ightarrow - 4^{-}}f(x)=\lim_{x
ightarrow - 4^{+}}f(x)\). Since \(f(x)=mx + b\) for \(x\in[-4,1]\), then \(\lim_{x
ightarrow - 4^{+}}f(x)=-4m + b\). So, \(-4m + b=9\).

Step3: Calculate the limit of the third - piece function

For \(x>1\), \(f(x)=5\cdot2^{1 - x}+34\). Calculate \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(5\cdot2^{1 - x}+34)=5\cdot2^{0}+34=39\).

Step4: Use continuity at \(x = 1\)

For the function to be continuous at \(x = 1\), \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)\). Since \(f(x)=mx + b\) for \(x\in[-4,1]\), then \(\lim_{x
ightarrow1^{-}}f(x)=m + b\). So, \(m + b=39\).

Step5: Solve the system of equations

We have the system of equations \(

$$\begin{cases}-4m + b=9\\m + b=39\end{cases}$$

\). Subtract the first equation from the second: \((m + b)-(-4m + b)=39 - 9\), which simplifies to \(5m=30\), so \(m = 6\).

Step6: Find the value of \(b\)

Substitute \(m = 6\) into \(m + b=39\), we get \(6 + b=39\), so \(b=33\).

Answer:

\(m = 6\), \(b = 33\)