Question

How do you show that `1 + 2x +x^3 + 4x^5 = 0` has exactly one real root?

Answer 1

See explanation.

Explanation:

First if we look at the limits of our function we see, that:

`lim_{x->-oo} f(x)=-oo`, and `lim_{x->+oo} f(x)=+oo`, so the function has at least one real root.

To show, that the function has only one real root we have to show, that it is monotonic in the whole `RR` To show it we have to calculate the derivative `f'(x)`

`f'(x)=20x^4+3x^2+2`

Now we have to solve: `20x^4+3x^2+2=0`

To do this we substitute `t=x^2` to transform the equation to a quadratic one:

`20t^2+3t+2=0`

This experssion is positive for all `x e RR`, because `Delta=-151<0`, so the function `f(x)` is increasing in the whole domain `RR`.

So we can conclude, that it has only one real root.

QED

Answer 2

See the explanation section.

Explanation:

Let `f(x) = 1+2x+x^3+4x^5` and note that for every `x`, `x` is a root of the equation if and only if `x` is a zero of `f`.

`f` has at least one real zero (and the equation has at least one real root).

`f` is a polynomial function, so it is continuous at every real number. In particular, `f` is continuous on the closed interval `[-1,0]`.

`f(-1) = 1-2-1-4 = -8` and `f(0)=1`

`0` is between `f(-1)` and `f(0`, so the Intermediate Value Theorem tells us that there is at least one number `c` in `(-1,0)` with `f(c) = 0`.

This `c` is a zero of `f` and a root of the equation.

`f` cannot have two (or more) zeros

Suppose that `f` had two (or more) zeros, call them `a` and `b`. So `f(a)=0=f(b)`

`f` is continuous on the closed interval `[a,b]` (it's still a polynomial) and

`f` is differentiable on the open interval `(a,b)`
(`f'(x) = 2+3x^2+20x^4` exists for all `x` in the interval.)

and `f(a) = f(b)` so by Rolle's Theorem (or by the Mean Value Theorem) there is a `c` in `(a,b)` with `f'(c)=0`

However,
`f'(x) = 2+3x^2+20x^4` can never be `0`.

(Look at each term. Both `20x^4` and `3x^2` are at least `0` and the constant adds `2`. So, `f'(x) >= 2`.)

That means that no `c` with `f'(c)=0` can exist.

Conclusion
If there were two (or more) zeros, then we could make `f'(c)=0`.
But, clearly, we cannot make `f'(c)=0`, so there cannot be two (or more) zeros.