What if the exponent in a power function is negative?

1 Answer
Dec 18, 2014

TLDR:
emathelp.net

Long version:

If the exponent of a power function is negative, you have two possibilities:

  • the exponent is even
  • the exponent is odd

The exponent is even:

#f(x) = x^(-n)# where #n# is even.
Anything to the negative power, means the reciprocal of the power.
This becomes #f(x) = 1/x^n#.
Now let's look at what happens to this function, when x is negative (left of the y-axis)
The denominator becomes positive, since you're multiplying a negative number by itself an even amount of time. The smaller#x# is (more to the left), the higher the denominator will get. The higher the denominator gets, the smaller the result gets (since dividing by a big number gives you a small number i.e. #1/1000#).

So to the left, the function value will be very close to the x-axis (very small) and positive.
The closer the number is to #0# (like -0.0001), the higher the function value will be. So the function increases (exponentially).

What happens at 0?

Well, let's fill it in in the function:

#1/x^n = 1/0^n#
#0^n# is still #0#. You're dividing by zero! ERROR, ERROR, ERROR!!
In mathematics, it is not allowed to divide by zero. We declare that the function doesn't exist at 0.
#x=0# is an asymptote.

What happens when x is positive?

When #x# is positive, #1/x^n#, stays positive, it will be an exact mirror image of the left side of the function. We say the function is even.

Putting it all together

Remember: we have established that the function is positive and increasing from the left side. That it doesn't exist when #x=0# and that the right side is a mirror image of the left side.

With these rules the function becomes:
enter image source here

What about an odd exponent?

The only change with an odd exponent, is that the left half becomes negative. It is mirrored horizontally. This function becomes:
enter image source here

Hope this helped!