Use Rolle's Theorem to show that between any two roots of ( x-1)sinx=o there exists at least one of tanx=1-x
If y = (x-1) sinx, then
y' = sinx + (x-1)cosx = k sin(x+θ)
where cosθ = 1/√(1+(x-1)^2)
since k is never zero, the zeroes of y' are the same as the zeroes of y, just shifted by θ. That is, they are between the zeroes of y
Note that if
sinx + (x-1) cosx = 0
tanx + (x-1) = 0 (for cosx≠0)
Student
Batiru
Maths.Assignment
Use Rolle'Theorm to show that between any to roots of(x_1)since=0 there exist at least one roots of tanx=1_x
To use Rolle's Theorem, let's start by understanding its conditions. According to Rolle's Theorem, if a function satisfies the following conditions:
1. The function is continuous on a closed interval [a, b],
2. The function is differentiable on the open interval (a, b), and
3. The function has the same value at both endpoints, f(a) = f(b),
Then, there exists at least one point c in the interval (a, b) where the derivative of the function is equal to zero, i.e., f'(c) = 0.
Now, let's apply Rolle's Theorem to the given problem. We need to find at least one root of the equation (x-1)sin(x) = 0 between any two roots of the equation tan(x) = 1 - x.
First, let's consider the equation (x - 1)sin(x) = 0. The roots of this equation are x = 0 and x = π (since sin(x) = 0 at these values).
Next, let's look at the equation tan(x) = 1 - x. Here, we notice that the range of values for x can be (-∞, ∞). However, to apply Rolle's Theorem, we need a closed interval. So, we need to determine a closed interval such that the function tan(x) is continuous on that interval.
Since the range of tan(x) is (-∞, ∞), we can choose the interval [-∞, ∞], which is a closed interval.
Now, let's check if the three conditions of Rolle's Theorem are satisfied in the given problem:
1. The interval [-∞, ∞] is a closed interval.
2. The function tan(x) is differentiable on the open interval (-∞, ∞) since the derivative of tan(x) is sec^2(x), and sec^2(x) is defined for all x except when cos(x) = 0 (i.e., at x = π/2 + nπ).
3. The function tan(x) satisfies f(-∞) = -∞ and f(∞) = ∞.
Since all three conditions are met, we can infer from Rolle's Theorem that there exists at least one point c in the interval (-∞, ∞) where the derivative of tan(x) is equal to zero, i.e., tan'(c) = 0.
Therefore, between any two roots of the equation (x - 1)sin(x) = 0, there exists at least one root of the equation tan(x) = 1 - x.