+145 Let \(g(x)\) be a function piecewise defined as \(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\)
If \(a\) is negative, find \(a\) so that \( g(g(g(10.5)))=g(g(g(a)))\)
I'm getting 10.5 but that can't be right. Also, I think the answer isn't a whole number.
+2440 DanielCai, obviously your answer of 10.5 cannot be right since \(a<0\). I would evaluate \(g(g(g(10.5)))\) one step at a time.
| \(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\) | This is the function's definition. Now, let's evaluate \(g(10.5)\) |
| \(g(10.5)=2\cdot 10.5-41\\ \hspace{13mm}=21-41\\ \hspace{13mm}=-20\) | Since \(10.5>0\), evaluate the bottom function. |
| \(g(-20)=-(-20)\\ \hspace{13mm}=20\) | Since \(-20\leq0\), evaluate the top function. |
| \(g(20)=2\cdot 20-41\\ \hspace{10mm}=40-41\\ \hspace{10mm}=-1\) | Since \(20>0\), evaluate the bottom function. |
Now, let's do the same process with \(g(g(g(a)))\).
| \(g(x) = \left\{ \begin{array}{cl} -x & x\le 0, \\ 2x-41 & x>0. \end{array} \right.\) | This is the function definition. | ||
| \(g(a)=-a\) | Since \(a\leq0\) was a set parameter given at the beginning of the problem, | ||
| \(g(-a)=-2a-41\) | Since multiplying a by -1 would make \(-a>0\), so evaluate it as the bottom fraction. | ||
| I do not know if \(-2a-41>0\), so I have to consider both cases. | ||
Set these two cases equal to -1 and check both solutions.
| \(2a+41=-1\) | \(-4a-123=-1\) |
| \(2a=-42\) | \(-4a=122\) |
| \(a=-21\) | \(a=-\frac{122}{4}=-\frac{61}{2}\) |
When checking both solutions, only \(a=-\frac{61}{2}\) works.
