by applying rolle's theorem, check whether it is possible that the function f(x) = x^5 + x -17 has two real roots. your reason is that if f(x) has two real roots then by rolle's theorem, f'(x) must be equal to what at certain value of x between these two roots, but f'(x) is always (negative, positive, zero)?
To check whether the function has two real roots using Rolle's theorem, we need to analyze the behavior of the derivative of the function, f'(x), to determine if there exists a point between the roots where the derivative is zero.
First, let's find the derivative of the given function, f(x) = x^5 + x - 17.
f'(x) = 5x^4 + 1
Now, we need to determine the sign of f'(x) to determine if it is always negative, positive, or zero.
To do this, we can observe that the derivative, f'(x) = 5x^4 + 1, is a polynomial of degree 4. We know that the derivative of a polynomial is also a polynomial, and since all polynomials are continuous, f'(x) is continuous.
To find where f'(x) is negative, positive, or zero, we need to solve for when f'(x) = 0. Let's set f'(x) = 0 and solve for x:
5x^4 + 1 = 0
5x^4 = -1
x^4 = -1/5
Here, we encounter a problem. The equation x^4 = -1/5 has no real solutions. This implies that f'(x) is never equal to zero, and therefore, never changes its sign.
According to Rolle's theorem, if f(x) has two real roots, then f'(x) must be zero at some point between the two roots. However, since f'(x) is always either negative or positive but never zero, it is not possible for the function f(x) = x^5 + x -17 to have two real roots by applying Rolle's theorem.