Proving the Maximum of a Cubic Polynomial: Solving for Cases of Equality

[.d9n.]

Homework Statement

Let f(x) = x^3+ ax^2+ bx + c with a, b, c real. Show that

M = max|f(x)|≥1/4
-1 ≤ x ≤ 1
and find all cases where equality occurs.

Homework Equations

The Attempt at a Solution

|(-1 ≤ x^3 ≤ 1)+(-a ≤ ax^2 ≤ a)+(-b ≤ bx ≤ b)+c|≥1/4?
not sure how to prove without knowing what a,b,c are?

 

[Mark44]

Homework Statement

Let f(x) = x^3+ ax^2+ bx + c with a, b, c real. Show that

M = max|f(x)|≥1/4
-1 ≤ x ≤ 1
and find all cases where equality occurs.

Homework Equations

The Attempt at a Solution

|(-1 ≤ x^3 ≤ 1)+(-a ≤ ax^2 ≤ a)+(-b ≤ bx ≤ b)+c|≥1/4?

Adding inequalities (which are either true or false) doesn't make any sense.

Basically, what you need to show is that the largest value of |f(x)| on the interval -1 <= x <= 1, is at least 1/4.

Wouldn't the derivative be useful here?

not sure how to prove without knowing what a,b,c are?

 

[.d9n.]

thanks
so i have
f '(x)=3x^2 + 2ax + b
f ' (x) =0 to find critical values
= -1/3a ± (√(4a^2 -12b))/6

not entirely sure where to go from here

 

[Mark44]

thanks
so i have
f '(x)=3x^2 + 2ax + b
f ' (x) =0 to find critical values
= -1/3a ± (√(4a^2 -12b))/6

1. When you start a line with =, you are saying that what is just above equals what's to the right of the = sign. In this case, you are saying that
0 = -1/3a ± (√(4a^2 -12b))/6
This is not true.
When you solved using the quadratic formula, what did you solve for?
2. There is an error what you did with the quadratic formula.

not entirely sure where to go from here

 

[.d9n.]

sorry i meant x= (-2a±√(4a^2-12b))/6,
not sure where I have gone wrong with the formula though (-b±√(b^2-4ac))/2a, correct?
although I am not too sure how this helps me

 

[.d9n.]

im not meant to do this:

(-1/3a ± (√(4a^2 -12b))/6)^3+ a(-1/3a ± (√(4a^2 -12b))/6)^2 +b(-1/3a ± (√(4a^2 -12b))/6)+c

surly?

 

[Mark44]

sorry i meant x= (-2a±√(4a^2-12b))/6,

That's better.

So what are the conditions for which f'(x) = 0? Will there be no such x-values, 1 such x-value, or 2 separate x-values?

not sure where I have gone wrong with the formula though (-b±√(b^2-4ac))/2a, correct?
although I am not too sure how this helps me

 

[Mark44]

im not meant to do this:

(-1/3a ± (√(4a^2 -12b))/6)^3+ a(-1/3a ± (√(4a^2 -12b))/6)^2 +b(-1/3a ± (√(4a^2 -12b))/6)+c

No

surly?

surly and surely are different words. Unless I miss my guess, you meant "surely."

 

[.d9n.]

yes i meant surely, sorry.
there will be two x values, assuming a,b≠0 i think

 

[.d9n.]

im unsure of how to work out what these x values are when we don't know what a and b are.
Is the second derivative useful once we have found out what the x values are to find the max
f''(x)=6x+2a

 

[Mark44]

yes i meant surely, sorry.
there will be two x values, assuming a,b≠0 i think

No, not true. If a = b = 2, there are no values for which f'(x) = 0.

Like I asked before, what are the conditions for which f'(x) = 0? For what values of a and be will there be no critical points, 1 critical point, or 2 separate critical points?

The Quadratic Formula gives you more information than you are using.

Also, all you are given is that a and b are real. You can't assume they aren't zero.

 

[Ray Vickson]

yes i meant surely, sorry.
there will be two x values, assuming a,b≠0 i think

The issue is not *just* whether there are two critical values of x, but whether or not these two values lie between -1 and +1. On the interval [-1,1], you may have: (i) no stationary point in (-1,1)(so the max/min occur at an endpoint x = +- 1); (ii) one stationary point in (-,1,1) and the other outside it (so the min occurs in (-1,1) and the max at +- 1, or vice-versa); (iii) two stationary points in (-1,1) (so that the max may occur in (-1,1) or at an endpoint, and similarly for the min.) Somehow you need to straighten out these possibilities.

RGV

 

[.d9n.]

the conditions are that the function has a maximum at x=n if f'(n)=0 and f''(0) is negative

it won't have any critical points when
-2a + √(4a^2 -12b)=0 so when a,b=0
it will have two critical point when
-2a - √(4a^2 -12b)=0 → b=2/3a^2 and a=√(3b/2)?

 

[.d9n.]

how am i meant to work out if the critical values lie in between -1 and 1, without know what a and b are. Sorry it all getting quite confusing

 

[Mark44]

the conditions are that the function has a maximum at x=n if f'(n)=0 and f''(0) is negative

it won't have any critical points when
-2a + √(4a^2 -12b)=0 so when a,b=0

Not quite, but you're getting warmer. There won't be any critical points when the discriminant < 0. When will there be one critical point? Two distinct critical points?

And as Ray pointed out, it makes a difference whether these points are in (-1, 1) or not.

it will have two critical point when
-2a - √(4a^2 -12b)=0 → b=2/3a^2 and a=√(3b/2)?

 

[.d9n.]

if the discriminant is > 0, the polynomial has two real roots, if the discriminant is = 0, the polynomial has one real root, and if the discriminant is < 0, the polynomial has no real roots
so therefore
if 4a^2-12b>0 it has two real roots so 0<4a^2-12b<=1,
if 4a^2-12b=0 it has one real root so 4a^2-12b=0
if 4a^2-12b<0 it has no real roots so -1<=4a^2-12b<0
is this what you mean?

 

[.d9n.]

when you say it makes a difference whether or not the points are in (-1,1) is this because its stated that x has to be in these points in the qu?
or is it something i have missed?
also what am i meant to do now, am i meant to solve for a and b?

 

[Mark44]

if the discriminant is > 0, the polynomial has two real roots, if the discriminant is = 0, the polynomial has one real root, and if the discriminant is < 0, the polynomial has no real roots
so therefore
if 4a^2-12b>0 it has two real roots so 0<4a^2-12b<=1,
if 4a^2-12b=0 it has one real root so 4a^2-12b=0
if 4a^2-12b<0 it has no real roots so -1<=4a^2-12b<0
is this what you mean?

Yes and no, but mostly no.

if 4a^2-12b>0 it has two real roots so 0<4a^2-12b<=1

What has two real roots? Be specific. Don't use the word "it".
Why do you think that 4a² - 12b <= 1? Where did that come from?

if 4a^2-12b=0 it has one real root so 4a^2-12b=0

if 4a^2-12b=0, then of course if 4a^2-12b=0.
Again, what has one real root?

if 4a^2-12b<0 it has no real roots so -1<=4a^2-12b<0

Why do you think that 4a² - 12b >= -1?

 

[.d9n.]

when i say it i mean the derivative of f'(x) or should i be working out the discriminant of f(x), i.e. b^2 c^2-4ac^3-4b^3d-27a^2d^2+18abcd
so in our case would be
a^2b^2-4b^3-4a^3-27c^2+18abc,

and i realize what wrong with the <=1 and >=1, i was thinking about x.

 

[.d9n.]

so therefore it would mean the polynomial

 

[SammyS]

so therefore it would mean the polynomial

Which polynomial?

There are two critical numbers (in addition to where f '(x)=0) for the interval [-1, 1].

They correspond to the end points of the interval. They are especially important if f '(x) has no roots on the interval [-1, 1] .

 

[Mark44]

when i say it i mean the derivative of f'(x) or should i be working out the discriminant of f(x), i.e. b^2 c^2-4ac^3-4b^3d-27a^2d^2+18abcd
so in our case would be
a^2b^2-4b^3-4a^3-27c^2+18abc,

and i realize what wrong with the <=1 and >=1, i was thinking about x.

No offense, but this problem seems way beyond your abilities. The idea of the discriminant in a quadratic equation does not extend to cubic polynomials, which is what f(x) is. I think that your best strategy might be to sit down with your instructor.