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

Question

Christopher Agresta · Accepted Answer

See the explanation section.

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.

Charles Anctil · Answer

See 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

Isaiah Allen · Answer

undefined To show that the equation $1 + 2x + x^3 + 4x^5 = 0$ has exactly one real root, we can use the intermediate value theorem. First, we observe that the polynomial is odd-degree, which means it will have at least one real root.

Next, we analyze the behavior of the polynomial as $x$ approaches negative infinity and positive infinity. As $x$ approaches negative infinity, the term with the highest power dominates, so the polynomial approaches negative infinity. As $x$ approaches positive infinity, the polynomial also approaches positive infinity.

Since the polynomial is continuous and changes sign from negative to positive, there must be at least one real root by the intermediate value theorem. To show there is exactly one real root, we can use calculus to find the derivative of the polynomial and show that it is always positive, indicating that the polynomial is strictly increasing and therefore can only intersect the x-axis once.