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.

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