The Student Room Group

How do I do this

19.png
Original post by KloppOClock
19.png


first equation rearranged gives x3+ax2−bx−c=0 \displaystyle x^3 +ax^2 - bx - c = 0

differentiating this and setting it equal to 0 to find stationary points; 3x2+2ax−b=0 \displaystyle 3x^2 +2ax - b = 0

quadratic formula and solving for x gives us two solutions, namely x1=−a+a2+3b3 \displaystyle x_{1} = \frac{-a + \sqrt{a^2 + 3b}}{3} and x2=−a−a2+3b3 \displaystyle x_2 = \frac{-a - \sqrt{a^2 + 3b}}{3} . Now since you are told all the numbers are positive x1>0 \displaystyle x_1 > 0 and x2<0 \displaystyle x_2 < 0 .

Differentiating again we get 6x+2a \displaystyle 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;

http://imgur.com/a/CsPoB

Do the exact same thing for the second equation and you'll sketch a graph which will have one negative and two positive roots
Original post by DylanJ42
first equation rearranged gives x3+ax2−bx−c=0 \displaystyle x^3 +ax^2 - bx - c = 0

differentiating this and setting it equal to 0 to find stationary points; 3x2+2ax−b=0 \displaystyle 3x^2 +2ax - b = 0

quadratic formula and solving for x gives us two solutions, namely x1=−a+a2+3b3 \displaystyle x_{1} = \frac{-a + \sqrt{a^2 + 3b}}{3} and x2=−a−a2+3b3 \displaystyle x_2 = \frac{-a - \sqrt{a^2 + 3b}}{3} . Now since you are told all the numbers are positive x1>0 \displaystyle x_1 > 0 and x2<0 \displaystyle x_2 < 0 .

Differentiating again we get 6x+2a \displaystyle 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;

http://imgur.com/a/CsPoB

Do the exact same thing for the second equation and you'll sketch a graph which will have one negative and two positive roots


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
Original post by KloppOClock
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


they cant be imaginary, for that to happen a2+3b<0 \displaystyle a^2 + 3b < 0 but you are told that b>0 on the top line
Original post by KloppOClock
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) \displaystyle 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=−a−a2+3b3 \displaystyle x_1 = \frac{-a-\sqrt{a^2 + 3b}}{3} . and x2=−a+a2+3b3 \displaystyle x_2 = \frac{-a+\sqrt{a^2 + 3b}}{3}

Now since you are told all the numbers are positive x1<0 \displaystyle x_1 < 0 and x2>0 \displaystyle x_2 > 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) \displaystyle f(x_2) is the minimum tp its less than -c which is less than 0, we can write f(x2)<−c<0 \displaystyle f(x_2) < -c < 0 .

Now for the third line to be true, f(x1)>0 \displaystyle f(x_1) > 0 , so writing this out fully we get;

(x1)3+a(x1)2−b(x1)−c>0 \displaystyle (x_1)^3 +a(x_1) ^2 - b(x_1) - c > 0

so to summarize we know x1<0 \displaystyle x_1 < 0 and from line 3 of the question know f(x1)>0 \displaystyle f(x_1) > 0 , we also know that the y intercept -c<0 \displaystyle \text{y intercept -c} < 0 , x2>0 \displaystyle x_2 > 0 and f(x2)<0 \displaystyle f(x_2) < 0 so the sketch of f(x) looks like so;

http://imgur.com/a/3kPfb

--------
--------

Okay now for the second equation,

ill call this equation g(x) \displaystyle g(x) for handiness

doing the same thing we get roots x3=a−a2+3b3 \displaystyle x_3 = \frac{a-\sqrt{a^2 + 3b}}{3} and x4=a+a2+3b3 \displaystyle x_4 = \frac{a+\sqrt{a^2 + 3b}}{3} . Once again since a,b,c > 0 this means x3<0 \displaystyle x_3 < 0 and x4>0 \displaystyle x_4 > 0

Again your happy that g(x3)>0 \displaystyle g(x_3) > 0 since its the maximum tp and must be >c>0 \displaystyle > c > 0 . We can write this as one inequality as g(x3)>c>0 \displaystyle g(x_3) > c > 0 .

Now we need to prove that g(x4)<0 \displaystyle g(x_4) < 0 , do you notice how x4=−x1 \displaystyle x_4 = -x_1 . So subbing −x1 \displaystyle -x_1 into g(x) yields;

g(x4)=g(−x1)=(−x1)3−a(−x1)2−b(−x1)+c \displaystyle g(x_4) = g(-x_1) = (-x_1)^3 - a(-x_1)^2 -b(-x_1) + c

g(x4)=−(x1)3−a(x1)2+b(x1)+c \displaystyle g(x_4) = -(x_1)^3 - a(x_1)^2 +b(x_1) + c

g(x4)=−[(x1)3+a(x1)2−b(x1)−c] \displaystyle g(x_4) = -[(x_1)^3 + a(x_1)^2 - b(x_1) - c]

Remember back to the first equation when line 3 told us that (x1)3+a(x1)2−b(x1)−c>0 \displaystyle (x_1)^3 +a(x_1) ^2 - b(x_1) - c > 0 therefore g(x4)=−[(x1)3+a(x1)2−b(x1)−c]<0 \displaystyle g(x_4) = -[(x_1)^3 + a(x_1)^2 - b(x_1) - c] < 0 just as we wanted.

So we know x3<0 \displaystyle x_3 < 0 , g(x3)>0 \displaystyle g(x_3) > 0 , the y intercept c>0 \displaystyle \text{y intercept c} > 0, x4>0 \displaystyle x_4 > 0 and that g(x4)<0 \displaystyle g(x_4) < 0 . Sketching this gives us;

http://imgur.com/a/2NInI

so g(x) \displaystyle 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)
wait till that mauritian priickk comes along and takes this down because it violates the rules smh
Original post by DylanJ42


Now we need to prove that g(x4)<0 \displaystyle g(x_4) < 0 , do you notice how x4=−x1 \displaystyle x_4 = -x_1 .

no

Original post by fksociety
wait till that mauritian priickk comes along and takes this down because it violates the rules smh

what?
Original post by KloppOClock
no


x1=−a−a2+3b3 \displaystyle x_1 = \frac{-a - \sqrt{a^2 + 3b}}{3} and x4=a+a2+3b3 \displaystyle x_4 = \frac{a + \sqrt{a^2 + 3b}}{3}

multiplying the left hand side one by -1 will give you the right hand side one

ie x1×−1=x4 \displaystyle x_1 \times -1 = x_4

Original post by KloppOClock
what?


hes referring to Zacken taking this down for full solution
Original post by DylanJ42
x1=−a−a2+3b3 \displaystyle x_1 = \frac{-a - \sqrt{a^2 + 3b}}{3} and x4=a+a2+3b3 \displaystyle x_4 = \frac{a + \sqrt{a^2 + 3b}}{3}

multiplying the left hand side one by -1 will give you the right hand side one

ie x1×−1=x4 \displaystyle x_1 \times -1 = x_4



hes referring to Zacken taking this down for full solution


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
Original post by KloppOClock
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 :dontknow:
Original post by KloppOClock
19.png


just realised how to do this much much quicker

call the first equationf(x) \displaystyle f(x) and say f(n)=n3+an2−bn−c=m \displaystyle f(n) = n^3 + an^2 - bn - c = m for some value of n

call the second equation g(x) \displaystyle g(x), now g(−n)=−n3−an2+bn+c=−[n3+an2−bn−c]=−m \displaystyle g(-n) = -n^3 -an^2 +bn + c = -[n^3 + an^2 - bn - c] = -m

so g(−x)=−f(x) \displaystyle g(-x) = -f(x)

therefore g(x) is just f(x) reflected the the x axis and the y axis (or vice versa)

do some sketches and get; http://imgur.com/a/LqM0S
Original post by DylanJ42
just realised how to do this much much quicker

call the first equationf(x) \displaystyle f(x) and say f(n)=n3+an2−bn−c=m \displaystyle f(n) = n^3 + an^2 - bn - c = m for some value of n

call the second equation g(x) \displaystyle g(x), now g(−n)=−n3−an2+bn+c=−[n3+an2−bn−c]=−m \displaystyle g(-n) = -n^3 -an^2 +bn + c = -[n^3 + an^2 - bn - c] = -m

so g(−x)=−f(x) \displaystyle g(-x) = -f(x)

therefore g(x) is just f(x) reflected the the x axis and the y axis (or vice versa)

do some sketches and get; http://imgur.com/a/LqM0S

but how would you know to sub -n into g(x)
Original post by KloppOClock
but how would you know to sub -n into g(x)


because of the differences in signs between the individual terms in f(x) and g(x).

Quick Reply

Latest