How do you find the constant a so that the function is continuous on the entire real line given f(x)=4, x <= -1, ax +b, -1< x <1 and 6, x>=1#?

Nov 6, 2016

We have $a = 1$ and $b = 5$ giving:

$f \left(x\right) = \left\{\begin{matrix}4 & x \le - 1 \\ x + 5 & - 1 \le x \le 1 \\ 6 & x \ge 1\end{matrix}\right.$

Explanation:

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

$f \left(x\right) = \left\{\begin{matrix}4 & x \le - 1 \\ a x + b & - 1 < x < 1 \\ 6 & x \ge 1\end{matrix}\right.$

Strictly speaking we want to find

$f \left(x\right) = \left\{\begin{matrix}4 & x \le - 1 \\ a x + b & - 1 \le x \le 1 \\ 6 & x \ge 1\end{matrix}\right.$

Just think about what we know so far, and how it would look: So for the mid interval we need a straight line passing through $\left(- 1 , 4\right)$ and $\left(1 , 6\right)$

This line would have the following gradient:
$m = \frac{\Delta y}{\Delta x} = \frac{6 - 4}{1 - \left(- 1\right)} = \frac{2}{2} = 1$
Hopefully you could also establish that by inspection!

So our required line passes through $\left(1 , 6\right)$ (equally we could you the other coordinate and get the same answer) and has gradient $m = 1$, so using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation is:

$y - 6 = \left(1\right) \left(x - 1\right)$
$\therefore y - 6 = x - 1$
$\therefore y = x + 5$

Which we can graph to confirm Hence, we have $a = 1$ and $b = 5$ giving:

$f \left(x\right) = \left\{\begin{matrix}4 & x \le - 1 \\ x + 5 & - 1 \le x \le 1 \\ 6 & x \ge 1\end{matrix}\right.$