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

Jul 27, 2019

#1
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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
+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.

hectictar Jul 27, 2019