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Let t(x) = sqrt{3x+1} and f(x)=5-t(x). What is t(f(5))?

 Feb 7, 2018

Best Answer 

 #1
avatar+2341 
+1

By the given information, we know the following:

 

\(t(x)=\sqrt{3x+1}\\ f(x)=5-t(x)\)

 

In order to find \(t(f(5))\), we must first evaluate f(x) when x=5:

 

\(f(x)=5-t(x)\) Since t(x) appears in the definition of f(x), plug in t(x) into the f(x) function.
\(f(x)=5-\sqrt{3x+1}\) Now, evaluate f(x) when x=5.
\(f(5)=5-\sqrt{3*5+1}\) Notice how every instance of x has been replaced wth a 5. Now, it is a matter of simplifying.
\(f(5)=5-\sqrt{16}\)  
\(f(5)=5-4\)  
\(f(5)=1\)  
   

 

Now we know that \(t(f(5)=t(1)\) since we just determined that \(f(5)=1\).

 

\(t(x)=\sqrt{3x+1}\) Of course, we want to evaluate when x=1, so replace every instance of x with a 1.
\(t(1)=\sqrt{3*1+1}\) Now, simplify.
\(t(1)=\sqrt{3+1}\)  
\(t(1)=\sqrt{4}=2\)  
   

 

Therefore, \(t(f(5))=2\)

.
 Feb 7, 2018
 #1
avatar+2341 
+1
Best Answer

By the given information, we know the following:

 

\(t(x)=\sqrt{3x+1}\\ f(x)=5-t(x)\)

 

In order to find \(t(f(5))\), we must first evaluate f(x) when x=5:

 

\(f(x)=5-t(x)\) Since t(x) appears in the definition of f(x), plug in t(x) into the f(x) function.
\(f(x)=5-\sqrt{3x+1}\) Now, evaluate f(x) when x=5.
\(f(5)=5-\sqrt{3*5+1}\) Notice how every instance of x has been replaced wth a 5. Now, it is a matter of simplifying.
\(f(5)=5-\sqrt{16}\)  
\(f(5)=5-4\)  
\(f(5)=1\)  
   

 

Now we know that \(t(f(5)=t(1)\) since we just determined that \(f(5)=1\).

 

\(t(x)=\sqrt{3x+1}\) Of course, we want to evaluate when x=1, so replace every instance of x with a 1.
\(t(1)=\sqrt{3*1+1}\) Now, simplify.
\(t(1)=\sqrt{3+1}\)  
\(t(1)=\sqrt{4}=2\)  
   

 

Therefore, \(t(f(5))=2\)

TheXSquaredFactor Feb 7, 2018

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