Prove that the equation 2x^5+x^3+3x+10=11 has exactly one solution

Question

Prove that the equation 2x⁵+x³+3x+10=11 has exactly one solution.

Answer 1

Rewrite as

2x^5 + x^3 + 3x -11 = 0

take the derivative and set it equal to zero to find local extrema

10x^4 + 3x^2 + 3 = 0, let y=x^, 10y^2 +3y + 3 =0

calculate the discriminate: b^2 -4ac = 9-4(10)(3) =- 111 <0

when the discriminate < 0, there are no real solutions, just 2 imaginary solutions

With a 5th degree polynomial, there are 5 solutions, here 4 are imaginary, only 1 is real

the one real solution is a proper fraction, between x = 0 and x = 1

f(0) = -1, f(1) = 5. somewhere inbetween 0 and 1, f(x) will equal zero. That's the only real solution

Answer 2

Because this is fifth degree polynomial, it can have AT MOST four turns. So let's examine where the turns might be. At any turn, the derivative must be 0.

If we start with the function and differentiate implicitly both sides:

2x^5+x^3+3x+10=11

d/dx (2x^5 + x^3 + 3x = 1)

10x^4 + 3x^2 + 3 = 0

By discovering where the x's are that make this equation true, we know where the curve turns flat.

However, because all the values on the left side of the equation must be positive for any x, the parts can never sum up to 0. So there are no solutions - which implies that the 5th degree polynomial has no turns so it therefore must cross the x-axis only once.