By applying Rolle's theorem, check whether it is possible that the function f(x)=x^5+x−5 has two real roots.
Answer: (input possible or impossible )
Your reason is that if f(x) has two real roots then by Rolle's theorem:
f′(x) must be (input a number here)
at certain value of x between these two roots, but f′(x) is always (input negative , positive, or zero )
To determine if the function f(x) = x^5 + x - 5 has two real roots by applying Rolle's theorem, we need to verify the conditions of the theorem.
Rolle's theorem states that if a function f(x) is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one value c in (a, b) such that f'(c) = 0.
Let's find the derivative of f(x) to determine f'(x):
f(x) = x^5 + x - 5
Taking the derivative with respect to x:
f'(x) = 5x^4 + 1
Now, let's analyze the sign of f'(x) for different values of x to determine if it is always negative, positive, or zero:
1. For x < 0:
- When we substitute x < 0 into f'(x) = 5x^4 + 1, all the terms are positive. Therefore, f'(x) is positive for x < 0.
2. For x = 0:
- Substituting x = 0 into f'(x) = 5x^4 + 1 gives f'(0) = 1. Therefore, f'(x) is positive at x = 0.
3. For x > 0:
- Similar to the case when x < 0, substituting x > 0 into f'(x) = 5x^4 + 1 gives positive terms. Thus, f'(x) is positive for x > 0.
Since f'(x) is always positive, it cannot be zero. Therefore, there is no value of x between the two roots of f(x) that satisfies the conditions of Rolle's theorem. Thus, it is impossible for the function f(x) = x^5 + x - 5 to have two real roots, according to Rolle's theorem.
f' = 5x^4+1
Looks like f' > 0 for all x.