+0

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

0
92
2

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

Guest Aug 29, 2017

#1
+18712
+2

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

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

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

$$f(f(f(f(f(1))))) = -33$$

heureka  Aug 29, 2017
Sort:

#1
+18712
+2

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

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

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

$$f(f(f(f(f(1))))) = -33$$

heureka  Aug 29, 2017
#2
+26322
+2

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

.

Alan  Aug 29, 2017

### 5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details