Roots of a cubic. Show that the function
f(x)=((x-a)^2) ((x-b)^2) + (x)
takes on the value ((a+b)/2) for some value of x.

Question

Roots of a cubic. Show that the function 
f(x)=((x-a)^2) ((x-b)^2) + (x)
takes on the value ((a+b)/2) for some value of x.

anonymous · Answer

i think we can use intermediate value theorem here...are u familiar with that?

anonymous · Answer

yes i am

anonymous · Answer

note that $f(a)=a$ ,  $f(b)=b$ and function is continous

anonymous · Answer

I get what you mean but how does it apply to this shoulwing that this has 3 solutions
and not just 1

anonymous · Answer

oops...i meant...I get what you said...but how do i show that this has 3 solutions and not just 1?

anonymous · Answer

this a polynomial from degree 4 !!

anonymous · Answer

ok that makes sense :)

anonymous · Answer

so it should have 4 solutions !!!

anonymous · Answer

the book says it has 3 solutions and we have to show that's true

anonymous · Answer

and how does that first equation take on the value (a+b)/2 for some value of x

anonymous · Answer

$f(a)=a$  and  $f(b)=b$  and we know that  $\frac{a+b}{2}$ lies on the interval [a,b] so according to the IVT there is a number c on interval [a,b] for which$$f(c)=\frac{a+b}{2}$$

anonymous · Answer

ohh okay but does the IVT only state that in between the a and b there lies only the value of 0 as one solution? I don't understand what the other two out of the three solutions are?

anonymous · Answer

to my understanding IVT has to do nothing with roots here

anonymous · Answer

Oh wait, I'm so sorry, I wrote the question entirely wrong. It's asking to show that the equation x^3 - 15x +1 = 0 has three solutions in the interval [-4,4]. 

Would this still require the IVT?

anonymous · Answer

oh :) that makes sense now

anonymous · Answer

yeah we can use IVT here

anonymous · Answer

$$f(x)=x^3-15x+1$$for example$$f(-4)=-64+60+1=-3$$$$f(-2)=-8+30+1=23$$so there is a root on interval [-4,-2] 

u try for intervals [-2,2]  and [2,4]

anonymous · Answer

i wish could give you another Medal mukushla :)

anonymous · Answer

oh tnx man

anonymous · Answer

ohhhhhhhh so just plug in and get the values

anonymous · Answer

yeah for interval [-4,-2]  f(-4)=-3  and f(-2)=23 since  -3<0<23 so accordind to IVT there is a 'c' on the interval such that f(c)=0

use IVT for other 2 intervals like this
good luck

anonymous · Answer

ohhhhhhh thank you soooooo much

anonymous · Answer

np :)

anonymous · Answer

nice i got it :)

anonymous · Answer

thank you again :) :D

anonymous · Answer

you are welcome :)