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Suppose $f(x)=x+1$. For what value of $x$ is $\overbrace{f(f(\cdots(f}^\text{2015 total \textit{f}s}(x))\cdots)=0$?

 Jul 19, 2019

Best Answer 

 #1
avatar+23137 
+2

Suppose \(f(x)=x+1\).
\(\text{For what value of $x$ is }\overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots)=0 ?\)

 

\(\begin{array}{|lrcll|} \hline f \\ \hline 1 & f(x) &=& x+1 \\ 2 & f\Big(f(x)\Big) = f(x+1) = (x+1)+1 &=& x+2 \\ 3 & f\Big(f\Big(f(x)\Big)\Big) = f(x+2) = (x+2)+1 &=& x+3 \\ \ldots \\ 2015 & \overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots) &=& x + 2015 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline x+2015 &=& 0 \\ \mathbf{x} &=& \mathbf{-2015} \\ \hline \end{array}\)

 

laugh

 Jul 19, 2019
 #1
avatar+23137 
+2
Best Answer

Suppose \(f(x)=x+1\).
\(\text{For what value of $x$ is }\overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots)=0 ?\)

 

\(\begin{array}{|lrcll|} \hline f \\ \hline 1 & f(x) &=& x+1 \\ 2 & f\Big(f(x)\Big) = f(x+1) = (x+1)+1 &=& x+2 \\ 3 & f\Big(f\Big(f(x)\Big)\Big) = f(x+2) = (x+2)+1 &=& x+3 \\ \ldots \\ 2015 & \overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots) &=& x + 2015 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline x+2015 &=& 0 \\ \mathbf{x} &=& \mathbf{-2015} \\ \hline \end{array}\)

 

laugh

heureka Jul 19, 2019

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