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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.

 Jun 27, 2018
 #1
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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. 
\(g(-2a-41)=-(-2a-41)\\ \hspace{23mm}=2a+41\) \(g(-2a-41)=2(-2a-41)-41\\ \hspace{24mm}=-4a-82-41\\ \hspace{24mm}=-4a-123\)

 

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. 

 Jun 27, 2018

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