Final answer:
Rolle's theorem is applicable to the polynomial function f(x)=x³+5x²-6x on the interval (0, 1). The function is continuous, differentiable, and equal at the endpoints; thus, there should be a point within the interval where the derivative equals zero.
Step-by-step explanation:
To verify the application of Rolle's theorem, we must first check if the function meets all the criteria required for the theorem to be applied. The two conditions that need to be satisfied are:
- The function f(x) is continuous on the closed interval [a, b].
- The function f(x) is differentiable on the open interval (a, b).
- The function f(x) has equal values at the endpoints, meaning f(a) = f(b).
In this case, the given function is f(x) = x³ + 5x² - 6x, and the interval is (0, 1). We first evaluate the function at the endpoints:
f(0) = 0³ + 5×0² - 6×0 = 0
f(1) = 1³ + 5×1² - 6×1 = 0
Both endpoints yield the same function value, zero, satisfying the third condition. Now we need to check for differentiability. As the function is a polynomial, it is differentiable everywhere on the real line, and hence, in the interval (0, 1).
Thus, Rolle's theorem is applicable to this function. According to the theorem, there should be at least one point c in the interval (0, 1) where the derivative f'(x) is zero.
To find the derivative, we use the power rule:
f'(x) = 3x² + 10x - 6
Now, set the derivative equal to zero and solve for x:
3x² + 10x - 6 = 0
This is a quadratic equation that can be solved using the quadratic formula or by factoring if possible. Solving for x provides the critical point(s) where the derivative equals zero. If such a point lies within the interval (0, 1), then Rolle's theorem is verified for this function within the given interval.