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Help on this problem

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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

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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