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Is \(f(x) = \frac{5^x - 1}{5^x + 1}\) an even function, odd function, or neither? 

 Jul 27, 2019

Best Answer 

 #1
avatar+8965 
+3

If     f( -x )  =  f(x)     then the function is even.

If     f( -x )  =  -f(x)    then the function is odd.

 

\(f(x) = \dfrac{5^x - 1}{5^x + 1}\)

                                       Plug in  -x  for  x

\(f(-x) = \dfrac{5^{(-x)} - 1}{5^{(-x)} + 1}\)

                                       Now we are looking for a way rewrite the right side so that  f(x)  appears.

                                       Let's rewrite  5(-x)  as  1 / 5x

\(f(-x) = \dfrac{\frac{1}{5^x} - 1}{\frac{1}{5^x} + 1}\)

                                             Multiply the numerator and denominator by  5x

\(f(-x) = \dfrac{\frac{1}{5^x} - 1}{\frac{1}{5^x} + 1}\cdot\dfrac{5^x}{5^x}\)

                                             Distribute the  5x  to the terms in the numerator and denominator

\(f(-x) = \dfrac{1 - 5^x}{1 + 5^x}\)

                                             Factor  -1  out of the numerator.

\(f(-x) = \dfrac{-1(-1 + 5^x)}{1 + 5^x}\)

                                             Addition can be done in any order so we can rearrange the terms like this..

\(f(-x) = \dfrac{-1( 5^x-1)}{ 5^x+1}\)

                                             Now we can write the  -1  beside the fraction like this...

\(f(-x) = -1\cdot\dfrac{ 5^x-1}{ 5^x+1}\)

                                             Finally, f(x)  has appeared!  \(f(x) = \dfrac{5^x - 1}{5^x + 1}\)   so we can substitute  f(x)  in for  \(\dfrac{5^x - 1}{5^x + 1}\)

\(f(-x) = -1\cdot f(x)\)

 

\(f(-x) = - f(x)\)

 

Remember that if     f( -x )  =  -f(x)    then the function is odd.

 

Since  f( -x )  =  -f(x)  , the function is odd.

 Jul 27, 2019
 #1
avatar+8965 
+3
Best Answer

If     f( -x )  =  f(x)     then the function is even.

If     f( -x )  =  -f(x)    then the function is odd.

 

\(f(x) = \dfrac{5^x - 1}{5^x + 1}\)

                                       Plug in  -x  for  x

\(f(-x) = \dfrac{5^{(-x)} - 1}{5^{(-x)} + 1}\)

                                       Now we are looking for a way rewrite the right side so that  f(x)  appears.

                                       Let's rewrite  5(-x)  as  1 / 5x

\(f(-x) = \dfrac{\frac{1}{5^x} - 1}{\frac{1}{5^x} + 1}\)

                                             Multiply the numerator and denominator by  5x

\(f(-x) = \dfrac{\frac{1}{5^x} - 1}{\frac{1}{5^x} + 1}\cdot\dfrac{5^x}{5^x}\)

                                             Distribute the  5x  to the terms in the numerator and denominator

\(f(-x) = \dfrac{1 - 5^x}{1 + 5^x}\)

                                             Factor  -1  out of the numerator.

\(f(-x) = \dfrac{-1(-1 + 5^x)}{1 + 5^x}\)

                                             Addition can be done in any order so we can rearrange the terms like this..

\(f(-x) = \dfrac{-1( 5^x-1)}{ 5^x+1}\)

                                             Now we can write the  -1  beside the fraction like this...

\(f(-x) = -1\cdot\dfrac{ 5^x-1}{ 5^x+1}\)

                                             Finally, f(x)  has appeared!  \(f(x) = \dfrac{5^x - 1}{5^x + 1}\)   so we can substitute  f(x)  in for  \(\dfrac{5^x - 1}{5^x + 1}\)

\(f(-x) = -1\cdot f(x)\)

 

\(f(-x) = - f(x)\)

 

Remember that if     f( -x )  =  -f(x)    then the function is odd.

 

Since  f( -x )  =  -f(x)  , the function is odd.

hectictar Jul 27, 2019

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