Prove that the function: f(x)=x^101+x^51+x+1, has neither a local maximum or minimum.

Question

I have tried to differentiate the function into f'(x)=101x¹⁰⁰+51x⁵⁰+1 and solve it to 0
101x¹⁰⁰+51x⁵⁰+1=0
101x¹⁰⁰+51x⁵⁰=-1
x⁵⁰(101x⁵⁰+51)=-1


Did i do it wrong .... Cause it seems it doesn't like the way how to solve the problem

Answer 1

That's an interesting way of solving the problem.

 

x⁵⁰ is always positive for all real values of x. 

This also means that 101x⁵⁰+51 is also always positive.

 

Therefore these 2 factors cannot multiply to be a negative number within the real numbers...so no max or mins.

 

When one exponent is double the other, you can make a u substitution and use the quadratic formula to find the actual values.

 

let u = x⁵⁰

 

This makes your equation 101u² + 51u +1 =0

 

So u = [-51 +- √(51²-4(101)(1) ] / 202 = [-51 +- √(2197) ] /202 

 

estimating these values you get the 2 answers -.020434818 and -.484515676

 

Now replace u to get x⁵⁰ = -.020434818....so on answer is +or- the 50th root o a negative number....which is complex.

Same goes for the other root....so no max or mins, but I've got the actual answers to the equation. ( note that if it was an odd root, you could take the odd root of a negative number, so the answers would not be complex)

 

 

Comes out the same in either case.

 

Hope this helped.