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Show that the equation dy/dx =0 has exactly 3 real solutions.y=(x-2)(x-3)(x-4)(x-5) +1

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jamnette | Student, Undergraduate | eNoter

Posted February 1, 2013 at 5:09 PM via web

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Show that the equation dy/dx =0 has exactly 3 real solutions.

y=(x-2)(x-3)(x-4)(x-5) +1

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted February 1, 2013 at 5:35 PM (Answer #1)

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First, since `y=f(x)=(x-2)(x-3)(x-4)(x-5)+1` is a fourth degree polynomial, its derivative is a third degree polynomial and can have at most three zeros. So if we can find three, then we know we got them all and we're done.

Since `f(x)` is continuous and differentiable everywhere, we can apply Rolle's Theorem on any interval. In particular, since `f(2)=f(3),` there is some point `c_1 in (2,3)` such that `f'(c_1)=0.` Using the exact same reasoning, we see that there must be points `c_2 in(3,4)` and `c_3 in(4,5)` where `f'(c_2)=f'(c_3)=0.` Note that `c_1,c_2,c_3` are all different points because they're in disjoint intervals, so we've found three zeros and know there can be no more.

Here's the graph:

 

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