How do you determine if #f(x) = |x-1|# is an even or odd function?

Question

Cora Alexander · Accepted Answer

Neither.

f(x) is even or odd according as #f(-x)=+-f(x)# #f(-x) =|-x-1|=|x+1|# and this is neither f(x) nor #f(-x)# .

Anna Allen · Answer

undefined To determine if $ f(x) = |x-1| $ is an even or odd function, we can use the definition of even and odd functions.

1. Even function: $ f(x) $ is even if $ f(-x) = f(x) $ for all $ x $ in the function's domain.
2. Odd function: $ f(x) $ is odd if $ f(-x) = -f(x) $ for all $ x $ in the function's domain.

For the function $ f(x) = |x-1| $, let's evaluate both $ f(-x) $ and $ f(x) $:

1. For $ f(-x) $:
   $$ f(-x) = |(-x)-1| = |-(x+1)| = |x+1| $$

2. For $ f(x) $:
   $$ f(x) = |x-1| $$

Now let's compare $ f(-x) $ and $ f(x) $:
$$ |x+1| \neq |x-1| $$

Since $ f(-x) $ is not equal to $ f(x) $, the function $ f(x) = |x-1| $ does not satisfy the condition for being even.

Next, let's check if it's odd by comparing $ f(-x) $ and $ -f(x) $:
$$ -f(x) = -|x-1| = -|-(x-1)| = -|1-x| $$

Since $ f(-x) $ is not equal to $ -f(x) $, the function $ f(x) = |x-1| $ does not satisfy the condition for being odd.

In conclusion, $ f(x) = |x-1| $ is neither an even nor an odd function.