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I was trying to find one similar to mine but couldn't find one that had 2 values for x and only 1 equation with an x.

$ f(x) = \begin{cases} 2, & x\le -1 \\ ax+b, & -1\lt x \lt 3 \\ -2, & x \ge 3 \\ \end{cases} $

I got a total of 4 equations when x is -1 and x is 3 for $a$ and $b$ respectively.

$b = 2 + a$ and $b = -2 - 3a$

$a = -2 + b$ and $a = \frac{-2-b}{3}$

Do I just set them equal to each other? Editted: I got $a = -1$ and $b = 1$ After setting the $b$ equations equal to each other and solving for a, and then setting the $a$ equations equal to each other and solving for b. Graphed it and it's correct.

Edit: Noticed that I had an equation wrong. Sign should be a plus not a negative to $b = 2 + a$.

Edit: Just graphed it to make sure it was correct; it is.

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    $\begingroup$ Your equation $b=2-a$ seems to have a sign wrong (it should be $b=2+a$) but your other equations are correct. It's worth noting there are only two distinct equations, and if you solve them you should get $(a, b) = (-1,1). $\endgroup$
    – jschnei
    Sep 8, 2015 at 6:13
  • $\begingroup$ How do yo get 4 equations? I just got two. At x = -1, ax + b = 2 and at x = 3, ax + b = -2. And your 2nd and 4th equations are same. $\endgroup$
    – user259381
    Sep 8, 2015 at 6:15
  • $\begingroup$ Thanks! I was wondering how it as 2 because I graphed it and it didn't make sense... I noticed that the graph is true when a is -1 so I realized there must be something wrong until you told me. Thanks! $\endgroup$
    – Hedylove
    Sep 8, 2015 at 6:15
  • $\begingroup$ I got 4 equations because you solve for a and b when x is -1. And you solve for a and b when x is 3. So you get 2 a = and 2 b= equations. $\endgroup$
    – Hedylove
    Sep 8, 2015 at 6:16
  • $\begingroup$ What you are saying as $4$ equations, that is actually $2$ equations. Last $2$ equations are the same. $\endgroup$
    – Rajat
    Sep 8, 2015 at 6:24

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