# Find the values of m and b that make f continuous everywhere: m = ? b = ?

## $f \left(x\right) = \left\{\left(\frac{7 + 6 x - {x}^{2}}{x + 1} \text{,", x < -1),(mx+b",", x ∈ [-1,4]),(2*2^(4-x)+16",} , x > 4\right)\right.$ Have never seen a piecewise function quite like this before and not sure how to even graph the mx+b line in all honesty.

Apr 16, 2017

$\left\{\begin{matrix}m = 2 \\ b = 10\end{matrix}\right.$

#### Explanation:

We can see that each individual function will be continuous on their domains.

To ensure that the function is continuous, we have to find the values of $m$ and $b$ that make the values of the functions equal at $x = - 1$ and $x = 4$, where the piecewise switches from one function to another.

First looking at $x = - 1$, we have to make $\frac{7 + 6 x - {x}^{2}}{x + 1}$ and $m x + b$ equal at $x = - 1$. If they have the same value, then the function is continuous there.

At $x = - 1$, we see that $\frac{7 + 6 x - {x}^{2}}{x + 1}$ is undefined, but we can find the limit at $- 1$ by factoring the numerator:

${\lim}_{x \rightarrow - 1} \frac{7 + 6 x - {x}^{2}}{x + 1} = {\lim}_{x \rightarrow - 1} \frac{\left(7 - x\right) \left(1 + x\right)}{x + 1} = {\lim}_{x \rightarrow - 1} \left(7 - x\right) = 8$

So, we need $m x + b$ to equal $8$ at $x = - 1$. That is, $- m + b = 8$.

We can't solve explicitly for $m$ or $b$ yet, so we also have to find a relation ensuring continuity at $x = 4$, where the function switches from $m x + b$ to $2 \cdot {2}^{4 - x} + 16$.

At $x = 4$, we see that $m x + b$ is equal to $4 m + b$ and the other function is $2 \cdot {2}^{0} + 16 = 18$. Then, $4 m + b = 18$.

So, we have the two relations of $m$ and $b$:

$\left\{\begin{matrix}- m + b = 8 \\ 4 m + b = 18\end{matrix}\right.$

Solving this gives:

$\left\{\begin{matrix}m = 2 \\ b = 10\end{matrix}\right.$