Is the following function odd, even, or neither? Give reasons for your answer.

f(x) = 1/(x-1)

The textbook answer section explains that it is neither. I assumed it would be odd because the power of x is 1 which is odd.

check to see whether

f(-x) = f(x). If so, it is even
So, check
1/(-x-1) = 1/(x-1)
it is clearly not even.

So, now check to see if it's odd: f(-x) = -f(x)
have at it.

f(x) = 1/(x-1)

it blows up when x = 1
a function f is even if f (-x) = f (x)
a function is odd if f (-x) = -f (x)
now try x = 2
f(-2) = f(2)???
1/(-2-1) = 1/(2-1)
1/-3 = 1/1 No, I think not odd then
but now
f(-2) = -f(2) ????
1/1 = - 1/1 No ! , Not even either

I have another question.

if a function has a larger exponent how does it affect its graph?

for example, if x^3 and x^6 are graphed. How can match each function to their graph.

The highest power affects the graph.

odd powers go off to infinity in both directions because odd powers of negative numbers are negative

even powers look more like parabolas, maybe with wiggles near the vertex, but since even powers are always positive, they go up on both ends.

To determine if a function is odd, even, or neither, we need to analyze its symmetry with respect to the y-axis or origin. In this case, we have the function:

f(x) = 1/(x-1)

To check for symmetry, we need to evaluate two key properties: f(-x) and -f(x).

1. Evaluating f(-x):
f(-x) = 1/(-x-1) = -1/(x+1)

2. Evaluating -f(x):
-f(x) = -1/(x-1)

Now, let's compare f(-x) and -f(x):
- f(-x) = -1/(x+1)
- -f(x) = -1/(x-1)

As you can see, f(-x) is not equal to -f(x). Therefore, the function does not satisfy the condition for odd symmetry.

For a function to be even, it needs to have symmetry with respect to the y-axis. That means if (x, y) is on the graph, then (-x, y) is also on the graph. As we can see, the function f(x) = 1/(x-1) does not exhibit this property.

Therefore, the function f(x) = 1/(x-1) is neither odd nor even.