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