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

Question

11-01-2019
Mathematics

contestada

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

Respuesta :

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LammettHash LammettHash · Answer 1 · 2019-01-11T19:03:33+01:00

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

Since [tex]a<0[/tex], we have [tex]g(a)=-a>0[/tex].

Since [tex]-a>0[/tex], [tex]g(g(a))=g(-a)=2(-a)-41=-2a-41[/tex].

Meanwhile, [tex]g(10.5)=-20[/tex], and [tex]g(g(10.5))=g(-20)=20[/tex], and [tex]g(g(g(10.5)))=g(g(-20))=g(20)=-1[/tex].

So we want to find [tex]a[/tex] such that [tex]g(-2a-41)=-1[/tex].

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

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

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

Respuesta :

Otras preguntas

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