differentiating this and setting it equal to 0 to find stationary points; 3x2+2ax−b=0
quadratic formula and solving for x gives us two solutions, namely x1​=3−a+a2+3b​​ and x2​=3−a−a2+3b​​. Now since you are told all the numbers are positive x1​>0 and x2​<0.
Differentiating again we get 6x+2a, its clear to see that a positive x value will > 0 and therefore minimum and therefore our negative x value is our maximum
Also since c > 0, and therefore -c < 0, we can sketch the first equation like so;
differentiating this and setting it equal to 0 to find stationary points; 3x2+2ax−b=0
quadratic formula and solving for x gives us two solutions, namely x1​=3−a+a2+3b​​ and x2​=3−a−a2+3b​​. Now since you are told all the numbers are positive x1​>0 and x2​<0.
Differentiating again we get 6x+2a, its clear to see that a positive x value will > 0 and therefore minimum and therefore our negative x value is our maximum
Also since c > 0, and therefore -c < 0, we can sketch the first equation like so;
Okay so you found the: y- intercept and the x values of the max and minimum
but how do you know if the roots are imaginary or real
I see what you mean now, my bad; (also i labelled the imgur picture wrong, x1 and x2 are mixed up, so ignore the last post and ill fix it up plus add in the extra bit to prove roots are real)
first equation rearranged gives , im going to call this f(x) for handiness
differentiating this and setting it equal to 0 to find stationary points;
quadratic formula and solving for x gives us two solutions, namely x1​=3−a−a2+3b​​. and x2​=3−a+a2+3b​​
Now since you are told all the numbers are positive x1​<0 and x2​>0
Differentiating again we get , its clear to see that a positive x value will > 0 and therefore minimum and therefore our negative x value is our maximum
Now the third line tells us that the three roots are real, its clear that since f(x2​) is the minimum tp its less than -c which is less than 0, we can write f(x2​)<−c<0.
Now for the third line to be true, f(x1​)>0, so writing this out fully we get;
(x1​)3+a(x1​)2−b(x1​)−c>0
so to summarize we know x1​<0 and from line 3 of the question know f(x1​)>0, we also know that the y intercept -c<0, x2​>0 and f(x2​)<0 so the sketch of f(x) looks like so;
Remember back to the first equation when line 3 told us that (x1​)3+a(x1​)2−b(x1​)−c>0 therefore g(x4​)=−[(x1​)3+a(x1​)2−b(x1​)−c]<0 just as we wanted.
So we know x3​<0, g(x3​)>0, the y intercept c>0, x4​>0 and that g(x4​)<0. Sketching this gives us;
so g(x) has a negative and 2 positive real roots, as required
sorry for this being so long, I didnt realise it involved so much working out (i originally thought this was a MAT multiple choice type question but obviously not)
damn i get it now, but that seems like a difficult question to get in under 3 minutes, thanks.
also, whats wrong with posting the full solution to a question and why would that anger zacken
yea there's probably a much easier way to do this, but that's my best attempt
its against forum rules to post full solutions and since Zacken got modded or whatever he had to tell people off for posting full solutions as they were breaking the rules. Others tell people off for it too but Zacken seems to be a favourite with the trolls so they pick on him