# How do we find the values of k and m that makes function continue anywhere piecewise function of (x^2) + 5 when x > 2, m(x+3) + k when -1 < x <=2 and 2(x^3) + x + 7 when x <=-1?

Nov 7, 2016

#### Answer:

$m = \frac{5}{3}$
$k = \frac{2}{3}$

#### Explanation:

Here is an example of a continuous function.

Here is an example of a discontinuous function.

So, continuous functions are where everything is connected; there are no gaps or holes.

The y-value of ${x}^{2} + 5$ when $x = 2$, (even though this would be a hole) is $9$.

We need the graph of $y = m \left(x + 3\right) + k$ to end at the point $\left(2 , 9\right)$.

The $y -$ value of $y = 2 {x}^{3} + x + 7$ at $x = - 1$ is $y = 2 {\left(- 1\right)}^{3} - 1 + 7 = - 2 - 1 + 7 = 4$.

We need the graph of $y = m \left(x + 3\right) + k$ to start at the point $\left(- 1 , 4\right)$.

Hence, we can write a systems of equations with respect to $m$ and $k$, since we know two inputs/outputs.

$4 = m \left(- 1 + 3\right) + k$
$9 = m \left(2 + 3\right) + k$

$k = 9 - 5 m$

$4 = 2 m + 9 - 5 m$

$- 5 = - 3 m$

$m = \frac{5}{3}$

$9 = 5 m + k$

$9 = 5 \left(\frac{5}{3}\right) + k$

$9 = \frac{25}{3} + k$

$k = \frac{2}{3}$

So, the function in the interval -1 < x ≤ 2 is $y = \frac{5}{3} \left(x + 3\right) + \frac{2}{3}$.

Nov 7, 2016

#### Answer:

$m = \frac{5}{3}$ and $k = \frac{2}{3}$,

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} + 5 & x > 2 \\ \frac{5}{3} \left(x + 3\right) + \frac{2}{3} & - 1 < x \le 2 \\ 2 {x}^{3} + x + 7 & x \le - 1\end{matrix}\right.$

#### Explanation:

We want to find $m$ and $k$ such that $f \left(x\right)$ is continuous:

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} + 5 & x > 2 \\ m \left(x + 3\right) + k & - 1 < x \le 2 \\ 2 {x}^{3} + x + 7 & x \le - 1\end{matrix}\right.$

Just think about what we know so far, and how it would look:

When $x = - 1 \implies f \left(- 1\right) = 2 {\left(- 1\right)}^{3} - 1 + 7 = - 2 - 1 + 7 = 4$
When $x = 2 \implies f \left(2\right) = {2}^{2} + 5 = 4 + 5 = 9$

So for the mid interval we need a straight line whose equation is $y = m \left(x + 3\right) + k$ which passes through $\left(- 1 , 4\right)$ and $\left(2 , 9\right)$

This line would have the following gradient:
$m = \frac{\Delta y}{\Delta x} = \frac{9 - 4}{2 - \left(- 1\right)} = \frac{5}{3}$

So our required line passes through $\left(- 1 , 4\right)$ (equally we could you the other coordinate and get the same answer) and has gradient $m = \frac{5}{3}$, so using $y = m \left(x + 3\right) + k$ we have:

$\frac{5}{3} \left(- 1 + 3\right) + k = 4$
$\frac{5}{3} \left(2\right) + k = 4$
$\frac{10}{3} + k = 4$
$k = 4 = \frac{10}{2} = \frac{2}{3}$

So the mid-section has equation $m \left(x + 3\right) = k$ with $m = \frac{5}{3}$ and $k = \frac{2}{3}$,

ie $y = \frac{5}{3} \left(x + 3\right) + \frac{2}{3}$,Which we can graph to confirm