Help, Please? (Question Down Below)

#P(x) = (x^4-2x^2-1)/(2x-1)#

What Is #P(1/2)#?

Thank You!

1 Answer
Advik S.
Mar 2, 2018

Input #1/2# instead of #x# in the function:

#(x^4-2x^2-1)/(2x-1)#

becomes:

#((1/2)^4-2*(1/2)^2-1)/(2*1/2-1)#

#=(1/8-(2*1/4)-1)/(1-1)#

#=(1/8-1/2-1)/(1-1)#

#=-1.375/0#

What do we have here!? We can't divide by zero!

As you can see, the function makes no sense. At the instantaneous point of #x=1/2#, the value of #y# will ascend to infinity, and is not seen on a graph.

This is what we call an asymptote. The limit of the function is not defined at the infinitesimally exact point of #x=1/2#.

If needed, use desmos.com to graph the function, and marvel at the amazing asymptote you get.

Related questions
Impact of this question
933 views around the world
You can reuse this answer
Creative Commons License