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For all real numbers x and y, the function f satisfies f(x + y) = f(x) + f(y).  If f(1) = 3, then find f(10).

 Dec 16, 2019
 #1
avatar+118 
+1

Since \(f(x+y)=f(x)+f(y)\), we can arrive at the following facts:

 

\(f(10)=f(2+8)=f(2)+f(8)\),

\(f(8)=f(2+6)=f(2)+f(6)\),

\(f(6)=f(2+4)=f(2)+f(4)\), and

\(f(4)=f(2+2)=f(2)+f(2)\).

Using these facts we can write

\(f(10)=f(2)+f(2)+f(2)+f(2)+f(2)\). In addition, since \(f(1)=3\), we can write

\(f(2)=f(1+1)=f(1)+f(1)=3+3=6 \). So

\(f(10)=6+6+6+6+6=30\).

 Dec 16, 2019
 #2
avatar+23809 
+2

For all real numbers x and y, the function f satisfies f(x + y) = f(x) + f(y).  If f(1) = 3, then find f(10).

 

\(\begin{array}{|lrcll|} \hline & \mathbf{f(x+y)} &=& \mathbf{f(x) + f(y)} \\\\ x=1,y=1: & f(1+1) &=& f(1) + f(1) \\ & \mathbf{f(2)} &=& \mathbf{2f(1)} \\\\ x=1,y=2: & f(1+2) &=& f(1) + f(2) \\ & f(3) &=& f(1) + 2f(1) \\ & \mathbf{f(3)} &=& \mathbf{3f(1)} \\\\ x=1,y=3: & f(1+3) &=& f(1) + f(3) \\ & f(4) &=& f(1) + 3f(1) \\ & \mathbf{f(4)} &=& \mathbf{4f(1)} \\\\ \ldots \\ & \mathbf{f(n)} &=& \mathbf{nf(1)} \\\\ \hline n=10: & f(10) &=& 10f(1) \quad | \quad f(1) =3 \\ & f(10) &=& 10*3 \\ & \mathbf{f(10)} &=& \mathbf{30} \\ \hline \end{array} \)

 

laugh

 Dec 17, 2019

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