The Second Derivative and Turning Points (1 Viewer)

[nick1048]

Prove that the graph of y = ax^3 + bx^2 + cx + d has two distinct turning points if b^2 > 3ac. Find the values of a, b, c and d for which a graph of this form has turning points at (1/2, 1) and (3/2, -1).

Seems to be a little sophisticated for a 2u question, nevertheless can anyone prove and find these points with working please. Your assistance is much appreciated.

 

[shafqat]

y' = 3a^2 + 2bx + c
now their are 2 turning pts if the derivative has two distinct solutions
for this, discriminant>0
so 4b^2-12ac>0
b^2>3ac

 

[shafqat]

for the rest, subsitute figures into y' and solve simultaneously
ill let u do that

 

[nick1048]

where did the 4b^2 - 12ac > 0 come from??? thanks for ur help btw.

for the rest, subsitute figures into y' and solve simultaneously
ill let u do that

I don't understand what you mean here.

 

[Xayma]

It's what is under the square root sign.

It is the discriminant designated by Δ

It indicates if there exists real roots and how many there are.

If &Delta;<0, there exists no real roots.
If &Delta;=0, there exists one real root.
If &Delta;>0, there exists two real roots.

 

[HellVeN]

discriminent = (b^2)-4ac

 

[nick1048]

ohhh and u just substitute the co-efficients in the problem right?
i.e. a becomes (3a), b becomes (2b) and c stays as c... it all makes sense, how delightful