Suppose $f(x)=x+1$. For what value of $x$ is $\overbrace{f(f(\cdots(f}^\text{2015 total \textit{f}s}(x))\cdots)=0$?
Suppose \(f(x)=x+1\).
\(\text{For what value of $x$ is }\overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots)=0 ?\)
\(\begin{array}{|lrcll|} \hline f \\ \hline 1 & f(x) &=& x+1 \\ 2 & f\Big(f(x)\Big) = f(x+1) = (x+1)+1 &=& x+2 \\ 3 & f\Big(f\Big(f(x)\Big)\Big) = f(x+2) = (x+2)+1 &=& x+3 \\ \ldots \\ 2015 & \overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots) &=& x + 2015 \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline x+2015 &=& 0 \\ \mathbf{x} &=& \mathbf{-2015} \\ \hline \end{array}\)
Suppose \(f(x)=x+1\).
\(\text{For what value of $x$ is }\overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots)=0 ?\)
\(\begin{array}{|lrcll|} \hline f \\ \hline 1 & f(x) &=& x+1 \\ 2 & f\Big(f(x)\Big) = f(x+1) = (x+1)+1 &=& x+2 \\ 3 & f\Big(f\Big(f(x)\Big)\Big) = f(x+2) = (x+2)+1 &=& x+3 \\ \ldots \\ 2015 & \overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots) &=& x + 2015 \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline x+2015 &=& 0 \\ \mathbf{x} &=& \mathbf{-2015} \\ \hline \end{array}\)