MHB Solving a Third-Degree Polynomial with Real Coefficients

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $f(x)$ be a third-degree polynomial with real coefficients satisfying $|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12$.

Find $|f(0)|$.
 
Mathematics news on Phys.org
Because a cubic polynomial shall have maximum 2 humps ( that is where it goes from down to up and from
up to down ) we shall have $f(1) = f(5)= f(6)$ and $f(2) = f(3) = f(7)$ and $f(1) \ne f(2)$.

For them to have values not meeting this criteria we shall have more humps .

As we are interested in absolute value we can choose $f(1) = - f(2) = 12$

As $f(1) = f(5)= f(6) = 12$ So we have $f(1) - 12 = f(5) - 12 = f(6) - 12 = 0$

So we can choose $f(x) = A(x-1)(x-5)(x-6) + 12 $ where A is a constant whose value we need to find

Putting $x=2$ we get $f(2) = A* 1 * (-3) *(-4) + 12 = 12A + 12 = - 12$ or A = -2

So $f(x) = -2(x-1)(x-5)(x-6) + 12$

Or $f(0) = -2 * (-1)(-5)(-6) + 12 = 72$

Or $\left | f(0) \right |= 72 $
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top