- #1
Jeff Ford
- 155
- 2
Prove [tex] x^4 + 4x + c = 0 [/tex] has at most two real roots
My thinking is that to prove this I would assume that it has three real roots and look for a contradiction.
So I set [tex] f(x) = x^4 + 4x + c [/tex] and assume three real roots [tex] x_1, x_2, x_3 [/tex] such that [tex] f(x_1) = f(x_2) = f(x_3) = 0 [/tex]
By MVT I know that there must exist [tex] c_1 [/tex] on the interval [tex] (x_1, x_2) [/tex] such that [tex] f'(c_1) = 0 [/tex] and [tex] c_2 [/tex] on the interval [tex] (x_2, x_3) [/tex] such that [tex] f'(c_2) = 0 [/tex] and [tex] c_3 [/tex] on the interval [tex] (x_1, x_3) [/tex] such that [tex] f'(c_3) = 0 [/tex]
Now I'm a little confused. Do I try to find three values of c that will satisfy [tex] f'(x) = 4x^3 + 4 = 0 [/tex]? Only the value x = -1 would make this true, so is that my contradiction? That MVT predicts three values and only one exists?
Thanks
Jeff
My thinking is that to prove this I would assume that it has three real roots and look for a contradiction.
So I set [tex] f(x) = x^4 + 4x + c [/tex] and assume three real roots [tex] x_1, x_2, x_3 [/tex] such that [tex] f(x_1) = f(x_2) = f(x_3) = 0 [/tex]
By MVT I know that there must exist [tex] c_1 [/tex] on the interval [tex] (x_1, x_2) [/tex] such that [tex] f'(c_1) = 0 [/tex] and [tex] c_2 [/tex] on the interval [tex] (x_2, x_3) [/tex] such that [tex] f'(c_2) = 0 [/tex] and [tex] c_3 [/tex] on the interval [tex] (x_1, x_3) [/tex] such that [tex] f'(c_3) = 0 [/tex]
Now I'm a little confused. Do I try to find three values of c that will satisfy [tex] f'(x) = 4x^3 + 4 = 0 [/tex]? Only the value x = -1 would make this true, so is that my contradiction? That MVT predicts three values and only one exists?
Thanks
Jeff
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