how shud i approach this q (1 Viewer)

[ekjchale#1]

 

[cossine]

Just use inspection. What happen as x -> 1 or infinity or 0.

Review limits if need be. There various definitions that form limits. You need to combine them.

e.g. standard definition, one sided, infinite limits, limits at infinity.

 

[yanujw]

1/f(x) has VAs where f(x) has zeroes (the converse is also true). The graph shown has asymptotes at 1, -1 and 0, so you can eliminate A. Then, test the sign of points at x=0.5 and x=1.5 to determine the equation.

Spoiler: Answer

B

 

[Deem_Skills]

1. The graph approaches infinity for both x and y (x-axis is asymptote)
2. The vertical asymptotes would be the x intersections of f(x), so just plug (-1, 0) and (1, 0) into the choices.
3. It will have a local maximum at where 1/f(x) has a local min.
That should be all you need, but the most important step is probably 2 as you can just plug in the points and determine one of A, B, C or D.

 

[Run hard@thehsc]

I feel like you can eyeball it to be B? (not sure if its the right ans). Basically, asymptotes form the x-int - and since the graph switches extremities as it passes through zero, one can deduce that it cannot be an x^2 int. So it could be B? (safe thing to sub in tho, but I feel eyeballing saves a lot of time)!!

 

[011235]

C has an x^2 and theres no turning point at x=0 so it can't be C

 

[Nezuko----]

Its B ( I am 90% sure) => you can kind of look off the graph, if the y value for (1/f(x)) is large then f(x) will be small vice versa, and keep going, of course there will be x intercepts on f(x) where there are x asymptotes on 1/f(x), but this is essentially to test whether or not you can identify these features.

 

[Run hard@thehsc]

C has an x^2 and theres no turning point at x=0 so it can't be C

Yh fair I meant B lol. DK why I said C, cos the thing does not bounce back up (as can be seen by the decreasing and increasing graphs within the asymptotes)

 

[Run hard@thehsc]

Its B ( I am 90% sure) => you can kind of look off the graph, if the y value for (1/f(x)) is large then f(x) will be small vice versa, and keep going, of course there will be x intercepts on f(x) where there are x asymptotes on 1/f(x), but this is essentially to test whether or not you can identify these features.

Yh it's def B

 

[Deem_Skills]

Yh fair I meant B lol. DK why I said C, cos the thing does not bounce back up (as can be seen by the decreasing and increasing graphs within the asymptotes)

yeah, also another way u could eliminate either B or C is by testing the sign on either side (that's usually what works best for me)

 

[jimmysmith560]

What would I need to answer for this?
Capture.PNG

I think that an answer to this question would be something along the lines of:

The graph of is a parabola with vertex . This fails the horizontal line test, so the inverse is not a function.

I hope this helps!