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Question:

pavelpimen
pavelpimen
Student
High School - 10th Grade

f(x)=(x-1)(x-3)(x-5)(x-7); f:R->R. Demonstrate that f'(x)=0 has exactly 3 real roots.

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Posted by pavelpimen on Saturday June 20, 2009 at 6:51 AM and tagged with derivative, math, real roots.

Answers:

  1. neela
    neela Teacher
    Graduate School

    eNotes Editor

    f(x)=(x-1)(x-3)(x-5)(x-7)

    We  differente both sides with respect to x. We establish that f'(7),f'(5),f'(3) and f'(1) have alternatiely different signs as below:

    f'(x)= (x-7)(x-5)(x-3)(x-1)[1/(x-7)+1/(x-5)+1/(x-3)+1/(x-1)]

    f'(7)=  (7-5)(7-3)(7-1)+ (7-7){...}= +ve + 0*{...)= +ve

    f'(5)=  (5-7)(5-3)(5-1)+(5-5){...} =(5-7)(+ve)+0{..}= -ve ,

    f'(3)=(3-7)(3-5)(3-1)+(3-3) {...} =(-ve)(-ve)(+ve)+0= +ve

    f(1)=(1-7)(1-5)(1-3)+(1-1){...)= (-ve)(-ve)(-ve)+0= -ve

    We, know, by Roll's theorem, that a continuous function g(x), if it changes its sign for x=a and x=b, then the function g(x) has to cross the x axis at a point x1 between x=a and x=b. Wherefore, g(x1) = 0,   (a<x1<b).

    By the above , we conclude f'(x) =0,   between 7  and  5;     5 and 3;   3 and  1. That is, f'(x) has 3 real roots.

     

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    Posted by neela on Saturday June 20, 2009 at 7:32 AM

  2. giorgiana1976
    giorgiana1976 Teacher
    Doctorate

    eNotes Editor

    Best answer as selected by question asker.

    In order to find out if the function has 3 real roots, we have to elaborate Rolle's queue, but for this we need to calculate the first derivative.

    For ease the calculus of the first derivative, we'll note x-4=y and we'll rewrite the function f(x)=(y+3)(y+1)(y-1)(y-3)=

    f(x)=y^4-10y^2+9=(x-4)^4-10(x-4)^2+9

    f'(x)=4(x-4)^3-20(x-4)=(x-4)(x^2-8x+11)

    f'(x)=0, (x-4)(x^2-8x+11)=0

    x-4=0,x=4

    x^2-8x+11=0

    x1=[ 8 +  sqrt(64-44)]/2=4+sqrt5

    x2=[8 - sqrt(64-44)]/2=4-sqrt5

    f(4)=(4-1)(4-3)(4-5)(4-7)=3*1*(-1)*(-3)=9

    f(4-sqrt5)=(4-sqrt5 -1)(4-sqrt5 -3)(4-sqrt5 -5)(4-sqrt5 -7)

    f(4-sqrt5)=(3-sqrt5)(1-sqrt5)(-1-sqrt5)(-3-sqrt5)

    sqrt5= approx 2,..

    f(4-sqrt5)= (0,..)(-)(-)(-)=-value

    4-sqrt5<4<4+sqrt5

    f(2)=(2-1)(2-3)(2-5)(2-7)=1*(-1)*(-3)*(-5)=-15

    f(1)=0

    1<4-sqrt5<4<4+sqrt5

    x      1  4-sqrt5  4  4+sqrt5

    f(x)   +     -       +      -

    We've noticed 3 changes of sign at f(x) values, that means that f(x) has 3 real roots in the following sets:

    x1 belongs to (1; 4 - sqrt5)

    x2 belongs to (4-sqrt5; 4)

    x3 belongs to (4; 4+sqrt5)

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    Posted by giorgiana1976 on Saturday June 20, 2009 at 10:14 AM