need help on graphing (1 Viewer)

[sonic1988]

Can some one plz help me to sketch f(x)=(x-1)/(x^3-x)

will f(x)=(x-1)/(x^3-x) and f(x)=1/(x^2+x) give u the same graph? In my opinion, i dont think so because each has different asymptotes. The first one has x=+/-1, 0 and the second one only has x=-1,0

 

[watatank]

Can some one plz help me to sketch f(x)=(x-1)/(x^3-x)

will f(x)=(x-1)/(x^3-x) and f(x)=1/(x^2+x) give u the same graph? In my opinion, i dont think so because each has different asymptotes. The first one has x=+/-1, 0 and the second one onlu has x=-1,0

to sketch f(x)=(x-1)/(x^3-x) =f(x)=(x-1)/(x(x^2-1)) = (x-1) / [x(x-1)(x+1)]

f(x) = (x-1) / [x(x-1)(x+1)] <leave it at that>

vertical assumptotes: where [x(x-1)(x+1)] = 0
x=0, +/-1

Horizontal assymptotes: as denominator gets insanely large the number gets smaller , so it tends to zero. so thats your hor. assymptote.

plot points at different x points to see how the graph will go.

 

[fishy89sg]

 

[zooy]

Well, good question, it will give a similar graph to f(x)=1/(x^2+x) but not the same. However, it is not about asymptote. In f(x)=(x-1)/(x^3-x), x cannot =1, so the function is discontinuous at the point (1, 1/2), however, in f(x) =1/(x^2+x), the curve will include the point (1, 1/2). Apart from this discontinuity, the two graphs are the same. Therefore, in order to plot f(x) = (x-1)/(x^3-x), you need to plot f(x) = 1/(x^2+x) first, where you have two asymptotes at x=0 and x=-1.(I assume you can plot the rest) And then u need to exclude the point (1,1/2)