4.2
18. show that the equation 2x-1-sinx=0 has exactly one real root.
sin x = 2x-1
2x = sin x - 1
range of sin x is -1 to + 1
so 2x is between -2 and 0
so x is between -1 and 0
To show that the equation 2x - 1 - sin(x) = 0 has exactly one real root, we need to use a combination of algebraic techniques and the properties of the given function.
Here's how to approach it:
Step 1: Graph the Function:
Begin by graphing the function y = 2x - 1 - sin(x) on a coordinate plane. This will help you visualize the behavior of the function and locate any possible real roots.
Step 2: Analyze the Behavior of the Function:
Note that the function y = 2x - 1 - sin(x) is continuous and differentiable over its entire domain. Additionally, since both 2x - 1 and sin(x) are continuous functions, their sum (2x - 1 - sin(x)) is also continuous. This means that the function doesn't have any "breaks" or gaps, and it can take on any real value.
Step 3: Use the Intermediate Value Theorem:
To prove that the equation has at least one real root, we can apply the Intermediate Value Theorem. This theorem states that if a continuous function has opposite signs for two points in its domain, then there must be at least one root between those two points.
In this case, consider the values of the function at x = 0 and x = 2. Calculate f(0) and f(2) as follows:
f(0) = 2(0) - 1 - sin(0) = -1,
f(2) = 2(2) - 1 - sin(2) ≈ 1.454.
Since f(0) is negative (less than zero) and f(2) is positive (greater than zero), by the Intermediate Value Theorem, there must be at least one real root between x = 0 and x = 2.
Step 4: Show that there is Only One Real Root:
To demonstrate that there is exactly one real root, we need to show that the equation doesn't have any additional roots in any other interval.
Since the graph of the function y = 2x - 1 - sin(x) is continuous and doesn't have any breaks or jumps, there cannot be any other root apart from the one we found between x = 0 and x = 2.
Therefore, we have proven that the equation 2x - 1 - sin(x) = 0 has exactly one real root.