Question

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 $g(g(g(10.5)))=g(g(g(a)))$.

Answer 1

`g(x)=\begin{cases}-x&\text{for }x\le0\\2x-41&\text{for }x>0\end{cases}`

Since
`a<0`, we have
`g(a)=-a>0`.

Since
`-a>0`,
`g(g(a))=g(-a)=2(-a)-41=-2a-41`.

Meanwhile,
`g(10.5)=-20`, and
`g(g(10.5))=g(-20)=20`, and
`g(g(g(10.5)))=g(g(-20))=g(20)=-1`.

So we want to find
`a` such that
`g(-2a-41)=-1`.

Suppose
`-2a-41\le0`, which happens if
`a\ge-20.5`. Then
`g(-2a-41)=2a+41`, so that
`2a+41=-1\implies a=-21`. But -21 is smaller than -20.5, so there's a contradiction.

This means we must have
`-2a-41>0`, which occurs for
`a<-20.5`. Then
`g(-2a-41)=2(-2a-41)-41=-4a-123`, so that
`-4a-123=-1\implies a=-30.5`.