Points of discountity

Question

how do we find the points of discountity of

f(x)=8x/5x^2-1?

 

If i made the denominator =0 i would get squared 1/5 and then when i will plug it back it will be undefined or am i doing it wrong !

 

Answer 1

the entire denominator should be enclosed in parentheses.

 

the denominator's zeros are x = ±1/√5; the correct say to express this is "plus or minus the reciprocal of the SQUARE ROOT of five."

 

These values are excluded from the domain of this function.  There are vertical asymptotes: x =  ±1/√5. That's the two discontinuities (note spelling).