you mean f'(x) has no real roots I think.
It does cross the x axis at least once as you showed with intermediate value
however the slope is never zero so it never can cross the axis again. The other four roots are therefore complex numbers.
Show that the equation x^5+x+1 = 0 has exactly one real root. Name the theorems you use to prove it.
I.V.T.
*f(x) is continuous
*Lim x-> inf x^5+x+1 = inf >0
*Lim x-> -inf x^5+x+1 = -inf <0
Rolles
*f(c)=f(d)=0
*f(x) is coninuous
*f(x) is differentiable
f'(x) = 5x^4+1=0
As f'(x) does not equal zero, because of rolles theorem, the assumptioin f(x) has two roots is false.
Used Intermediate Value theorem and Rolles Theorm.
1 answer
1 answer