Question

In Exercises 49 and \(50,\) use the Intermediate Value Theorem and Rolle's Theorem to prove that the equation has exactly one real Solution. $$ x^{5}+x^{3}+x+1=0 $$

Short answer

By applying the Intermediate Value Theorem and Rolle's Theorem on the function \(f(x) = x^{5}+x^{3}+x+1\), it can be proven that the given equation has exactly one real root.

Step-by-step solution

Define the Function and Check Polynomial Properties

Firstly, define the function \(f(x)\) as \(f(x) = x^{5}+x^{3}+x+1.\) It is crucial to note that this is a polynomial function, which implies that it is both continuous and differentiable at every value of \(x\) in the real number line.

Apply the Intermediate Value Theorem

Choose two values of \(x\) which are of opposite signs and will help to demonstrate if there is a real root. By plugging -2 into the function, we get \( f(-2) = -30\), and by plugging in 0 into the function, we get \( f(0) = 1\). Given that \(f(x)\) is continuous for all \(x\), and \(f(-2) < 0\) while \(f(0) > 0\), by the Intermediate Value Theorem there must be a point \(c\) in the interval \(-2 < c < 0\) at which \(f(c) = 0\). It implies that there is at least one real root in this interval.

Apply Rolle's Theorem to Show Singularity of the Root

Now, take the derivative of the function \(f'(x) = 5x^{4}+3x^{2}+1\). We can see easily that \(f'(x) > 0\) for all \(x\) in the real number line because \(5x^{4}+3x^{2}\) are always non-negative (equal or greater than 0) and adding 1 makes it strictly positive. The derivative of a function signifies its slope. Therefore, the function is always increasing, which suggests the function \(f(x)\) can cross the x-axis exactly once. This is supported by Rolle's Theorem that there must be at exactly one stationary point between any two points if the function value at two points are equal. In conclusion, the equation \(f(x)=0\) has precisely one real root.