Question

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

Answer 1

Hint: Given a polynomial. We have to prove that the polynomial has exactly one real root. First, we will determine the interval in which the polynomial is equal to zero using the intermediate value theorem. Then, we will assume that the function has two or more zeros. Then differentiate the polynomial in some open interval. Then, apply Rolle’s theorem to find the value at which differentiation of the function is equal to zero. If $f'\left( x \right) \ne 0$, then two or more zeros are not possible.

Complete step by step solution:
We are given the polynomial, $1 + 2x + {x^3} + 4{x^5} = 0$.

Since the function is a polynomial, therefore it is continuous at every real number.

Now, we will check whether the polynomial is continuous on the closed interval $\left[ { - 1,0} \right]$

We will substitute $ - 1$ for $x$ into the polynomial.

 $ \Rightarrow f\left( { - 1} \right) = 1 + 2\left( { - 1} \right) + {\left( { - 1} \right)^3} + 4{\left( { - 1} \right)^5}$

Simplify the expression, we get:

$ \Rightarrow f\left( { - 1} \right) = 1 - 2 - 1 - 4 = - 6$

We will substitute $0$ for $x$ into the polynomial.

 $ \Rightarrow f\left( 0 \right) = 1 + 2\left( 0 \right) + {\left( 0 \right)^3} + 4{\left( 0 \right)^5}$

Simplify the expression, we get:

$ \Rightarrow f\left( 0 \right) = 1 + 0 + 0 + 0 = 1$

0 lies between $f\left( { - 1} \right)$ and $f\left( 0 \right)$. Therefore, according to intermediate value theorem there must exist one number k in the open interval $\left( { - 1,0} \right)$for which $f\left( k \right) = 0$

Thus, k is the root of the polynomial.

Now, we will assume that the function has two or more zeros, $a$ and $b$ such that $f\left( a \right) = 0 = f\left( b \right)$

Now, the function is continuous on the closed interval $\left[ {a,b} \right]$ and also it is differentiable on the open interval $\left( {a,b} \right)$

Now, we will differentiate the function.

$ \Rightarrow \dfrac{d}{{dx}}\left( {1 + 2x + {x^3} + 4{x^5}} \right)$

Now, apply the chain rule of differentiation.

$ \Rightarrow \dfrac{d}{{dx}}\left( {1 + 2x + {x^3} + 4{x^5}} \right) = 0 + 2 + 3{x^2} + 4 \times 5{x^4}$

$ \Rightarrow \dfrac{d}{{dx}}\left( {1 + 2x + {x^3} + 4{x^5}} \right) = 2 + 3{x^2} + 20{x^4}$

Now, apply the Rolle’s theorem in which there must exist a number k in the interval $\left( {a,b} \right)$ for which $f'\left( k \right) = 0$

Here, the polynomial $ \Rightarrow f'\left( x \right) = 2 + 3{x^2} + 20{x^4}$ can never be zero.

Therefore, $f'\left( x \right) > 0$ which means there is no k that exists.

Thus, our assumption contradicts here, and there is only one real root of the function.

Hence the polynomial $1 + 2x + {x^3} + 4{x^5} = 0$ has exactly one real root.

Note: The students please note that the real roots of the polynomial must exist because the function is continuous on every point in the interval. Students must remember that we have applied intermediate value theorem which states that for a given function $f\left( x \right)$continuous on interval $\left[ {a,b} \right]$ then there must exists c such that $f\left( b \right) < f\left( c \right) < f\left( a \right)$.
We have also applied Rolle’s theorem which states that if a function is continuous on interval $\left[ {a,b} \right]$ and differentiable on $\left( {a,b} \right)$ where $f\left( a \right) = f\left( b \right)$, then there must exists c such that $f'\left( c \right) = 0$