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by **nona.m.nona** on Thu Jan 22, 2009 5:29 pm

Prove that the function f(x) = x^101 + x^51 + x + 1 has neither a local maximum nor a local minimum.

I can take the derivative: f'(x) = 101x^100 + 51x^50 + 1

But what then? :?:

**Sponsor**

by **stapel_eliz** on Thu Jan 22, 2009 5:35 pm

Use what you learned back in algebra. For f(x) to have a local max or min, there has to be a critical point, right? So you'd have to be able to find a zero for the derivative. This derivative is a quadratic in x⁵⁰:

. . . . . $101 x^{100}\, +\, 51 x^{50}\, +\, 1\, =\, 101\left(x^{50}\right)^2\, +\, 51\left(x^{50}\right)\, +\, 1$

Set the right-hand side above equal to zero, and apply the Quadratic Formula to find the solution value for x⁵⁰. Can you then find a real value for x (being an even-index root of x⁵⁰)? :wink:

Eliz.

by **nona.m.nona** on Mon Jan 26, 2009 1:08 pm

The quadratic formula gives me x^50 = (-51+/-sqrt[(-51)^2-4(101)])/(2*101) = (-51+/-sqrt[2197])/(202) = (-51+/-46.9)/(202). The -51+/-46.9 will always be negative, so I can't do the 50th root of it to get a number for x. So this means I can't get a zero, so there isn't a critical point, so there can't be a max or min, right?

by **stapel_eliz** on Mon Jan 26, 2009 1:29 pm

nona.m.nona wrote:...The -51+/-46.9 will always be negative, so I can't do the 50th root of it to get a number for x. So this means I can't get a zero, so there isn't a critical point, so there can't be a max or min, right?

Exactly right!

Eliz.

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Prove f(x) = x^101 + x^51 + x + 1 has no local max/min

Prove f(x) = x^101 + x^51 + x + 1 has no local max/min

Re: Prove f(x) = x^101 + x^51 + x + 1 has no local max/min