Does y = ax^3 + bx^2 + cx + d Have Two Distinct Turning Points if b^2 > 3ac?

[disfused_3289]

Prove that the graph of y= ax^3 + bx^2 + cx + d has two distinct turning points if
b^2> 3ac. Find values of a,b,c and d for which the graph of this form has turning points at (0.5, 1) and (1.5, -1)

 

[sadhu]

hi iam a new member to this site

well here isolved first part of it
differentiate the whole equation with respeact to x

you will get

3ax^2+2bx+c

find the maxima and minima

i.e for two turning points two distinct max,min must exist

Discriminant>0

(2b)^2-12ac>0
4*b^2 > 12 ac
b^2 >3 ac...prooved

 

[sadhu]

second part

3ax^2+2bx+c=0 was the equation i already mentioned
one answer was 1/2,second 3/2
adding them 2
multiplying them we get 3/4

2=-2b/3a
-b=3a

3/4=c/3a
4c=9a

substituting value of function in the function equation we get

a+2b+4c+8d=8
27a+18b+12c+8d=-8

solve these equations simultaniously
for a,b,c,d

we get

d=-1
a=4
b=-12
c=9

if iam correct

 

[mathwonk]

i thought saddhu had 2 dd's. aha the free dictionaire gives both variants. you learn something every day, ram, ram.

http://www.flickr.com/photos/laurentis/1776669597/

 

[sadhu]

I don't understand your last statement ..
please pardon...

 

[disfused_3289]

Solved

Ah, I now understand it. I never thought of delta before
Thanks alot