Can we find at most two real roots for the equation x^4 + 4x + c = 0?

[helpm3pl3ase]

This theorem is confusing me even though it is sittin right in front of me.. I am given an equation x^4 + 4x + c = 0 and asked to find at most two real roots??

I know we need to take the derivative, but from there I am lost.

 

[courtrigrad]

If $$f(x)$$ is differentiable in the open interval $$(a,b)$$ and continuous on the closed interval $$[a,b]$$, then there is at least one point $$c$$ in $$(a,b)$$ such that:

$$f'(c) = \frac{f(b)-f(a)}{b-a}$$

Assume that there are two real roots $$c_{1}$$ and $$c_{2}$$ where $$c_{1} < c_{2}$$.Then $$f(c_{1}) = 0 = f(c_{2})$$.

Thus $$4x^{3} + 4 = 0$$

 

[mathwonk]

basic principle of garphing: a graph can only change direction at a critical pont, and not always then.

 

[lord12]

x^(4) + 4x + c = 0
The function is a polynomial and is differentiable and continuous. Suppose a and b are distinct roots. There exists a c in which a<c<b such that 0 = f(b) - f(a). Since f'(x)= 4x^(2) + 4>0, f(a) != f(b). This is a contradiction; hence, a and b cannot both be roots.