Submitted by Anonymous on Mon, 10/17/2016 - 10:06
Forums:
Assignment 9 is now available.
I have also inserted links to websites where you can read further about some of the topics and ideas introduced in the assignments for all the previous assignments, so you may like to revisit them!
Q3
Should the last part not be 27b^2 + 4ba^3 ≥ 0 instead of >0 because fewer than 3 roots includes 2 which would occur when one of the turning points is at y=0, eg: if b=0 there are 2 solutions x=0 or x=-a, but 27b^2 + 4ba^3 = 0
If it is true for 27b^2+4ba^3
If it is true for 27b^2+4ba^3 ≥ 0 then it is certainly true for 27b^2 + 4ba^3>0, but they've only asked you to show the weaker condition, perhaps because they want you to consider the case when you have equality in order to determine the strict inequality for the case where it's <0 and there are three distinct roots. Don't worry too much if a STEP question asks you to prove something and you can actually prove a stronger condition!
Q3
I understand that for the sketches, A > 0 B > 0 and A < 0 B > 0, share a turning at x = 0 but I'm unsure why they share the other turning point of (2/3 A) because surely when A < 0 , it has a turning point at 2/3 A and not - 2/3 A?
Q3
If $A$ is a negative number then $-\frac{2}{3}A$ will be positive, which is what you want.
Q3
*why they share the other turning point of - 2/3 A (not 2/3A)
Q3
In both cases the turning points are at $x=0$ and $x=-\frac 2 3 A$.
If $A>0$ then this second turning point will be for a negative value of $x$ (and will be a maximum) where as if $A\lt 0$ then $-\frac 2 3 A$ is positive and this turning point is at a positive value of $x$ (and is a minimum).
Perhaps try using https://www.desmos.com/ to sketch some curves and see what is happening. You can ask it to sketch $y=x^3+Ax^2 + B$ and add some "sliders" which you can slide to see what happens as $A$ and $B$ change. You can also plot $x=-\frac 2 3 A$ to help convince you that the second turning point always has $x$ coordinate $-\frac 2 3 A$.