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

2 Answers
Sep 29, 2015

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

Sep 29, 2015

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.