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Let \[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \]Compute $f(f(f(f(f(1))))).$

 Aug 29, 2017

Best Answer 

 #1
avatar+22182 
+1

Let \[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \]Compute $f(f(f(f(f(1))))).$

 

\(\text{Let } \left[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \right] \\ \text{Compute } f(f(f(f(f(1))))). \)

 

\(\begin{array}{|llcll|} \hline x=1 \quad & f(1) = -1^2 &=& -1 & | \quad (x\ge 0) \\ x=-1 \quad & f(-1) = -1+8 &=& 7 & | \quad (x \lt 0) \\ x=7 \quad & f(7) = -7^2 &=& -49 & | \quad (x\ge 0) \\ x=-49 \quad & f(-49) = -49+8 &=& -41 & | \quad (x \lt 0) \\ x=-41 \quad & f(-41) = -41+8 &=& -33 & | \quad (x \lt 0) \\ \hline \end{array} \)

 

\(f(f(f(f(f(1))))) = -33\)

 

laugh

 Aug 29, 2017
 #1
avatar+22182 
+1
Best Answer

Let \[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \]Compute $f(f(f(f(f(1))))).$

 

\(\text{Let } \left[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \right] \\ \text{Compute } f(f(f(f(f(1))))). \)

 

\(\begin{array}{|llcll|} \hline x=1 \quad & f(1) = -1^2 &=& -1 & | \quad (x\ge 0) \\ x=-1 \quad & f(-1) = -1+8 &=& 7 & | \quad (x \lt 0) \\ x=7 \quad & f(7) = -7^2 &=& -49 & | \quad (x\ge 0) \\ x=-49 \quad & f(-49) = -49+8 &=& -41 & | \quad (x \lt 0) \\ x=-41 \quad & f(-41) = -41+8 &=& -33 & | \quad (x \lt 0) \\ \hline \end{array} \)

 

\(f(f(f(f(f(1))))) = -33\)

 

laugh

heureka Aug 29, 2017
 #2
avatar+27837 
+1

Interestingly, a staircase diagram shows this function is cyclic once it reaches x = -1:

 

.

 Aug 29, 2017

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