Hsc 2012 q8 (1 Viewer)

[AnimeX]

Hi can someone explain this MC to me please?

Having a bit of trouble, how do you manipulate it? so do you think of the graph as a cubic and integrate it so thus it's a quartic now?

then what :s?

Also, if f(x) = x - 4/x

how would you do e^f(x) and ln(f(x)), when I tried them I got completely the wrong answer.

 

[nightweaver066]

Q8. By looking at the graph, you can tell there's a horizontal inflexion at x = 1 and a minimum turning point at x = -5/4.

Now you know that the equation must have (x-1)^3 in it.

If you do a quick sketch of the possible graph, you notice how there must be an x-intercept less than -5/4, so (x+2) must also be apart of it.

So your answer is B.

Other question:
Try following this:
http://i.imgur.com/JCIQGq9.png

You basically just draw f(x) = x - 4/x first, then look at the various parts of the graph for specific features and draw them in.

 

[AnimeX]

Q8. By looking at the graph, you can tell there's a horizontal inflexion at x = 1 and a minimum turning point at x = -5/4.

Now you know that the equation must have (x-1)^3 in it.

If you do a quick sketch of the possible graph, you notice how there must be an x-intercept less than -5/4, so (x+2) must also be apart of it.

So your answer is B.

Other question:
Try following this:
http://i.imgur.com/JCIQGq9.png

You basically just draw f(x) = x - 4/x first, then look at the various parts of the graph for specific features and draw them in.

thanks , the min t.p for the multiple choice wouldn't be on y=0 right? how do you know this?

 

[nightweaver066]

thanks , the min t.p for the multiple choice wouldn't be on y=0 right? how do you know this?

It could. The minimum turning point can have any y-value as the graph of y = P(x) can be shifted in any way vertically by a constant (which is annihilated after differentiation).