prove that 5x - 7 - sin3x = 0 has at least one zero and prove that it has exactly one real zero.
How am I supposed to show my work for this? I don't really understand how to show my work for the IVT. Without that I can't do any work for the MVT.
Draw graphs of 5x - 7 and sin3x and see how many times they intersect. They only cross once, at about x = 1.27. That can be clearly seen from the two graphs.
I believe in doing things the easy way. I don't see how either the mean or intermediate value theorems will be useful here, but math is not my specialty.
5x - 7 only has values that are in the +/- 1 range of sin 3x when x is between 1.2 and 1.6. Perhaps that will help with a more formal proof.
You only have to consider the function
5x - 7 - sin3x between x = 1.2 and x = 1.6.
In that interval, it goes from -0.5575 to 1.996.
The slope remains positive so it can only equal zero once.
To prove that the equation 5x - 7 - sin(3x) = 0 has at least one zero, and to further prove that it has exactly one real zero, you can use the Intermediate Value Theorem (IVT) and the Mean Value Theorem (MVT), respectively.
1. Proving at least one zero using the IVT:
To apply the IVT, you need to show that the function is continuous over a given interval and that it takes on positive and negative values at the endpoints of that interval.
In this case, consider the function f(x) = 5x - 7 - sin(3x). To demonstrate continuity, note that 5x and sin(3x) are both continuous functions over the entire real number line. Therefore, their sum, 5x - 7 - sin(3x), is also continuous over the same interval.
To apply the IVT, you need to find two values of x, a and b, such that f(a) < 0 and f(b) > 0. By analyzing the behavior of the function or graphing it, you can make an educated guess to approximate its roots. Then, you can use numerical methods such as a graphing calculator, a math software, or an online tool to get approximate values for these roots.
For example, by sketching the graph of f(x) = 5x - 7 - sin(3x) or using a calculator, you may find that f(0) < 0 and f(1) > 0. This means that the function changes sign between x = 0 and x = 1, suggesting the existence of at least one root in this interval, based on the IVT.
2. Proving exactly one real zero using the MVT:
To prove that there is exactly one real zero, you can analyze the derivative of the function.
Let's find the derivative of f(x) = 5x - 7 - sin(3x):
f'(x) = 5 - 3cos(3x).
Note that the derivative f'(x) is also continuous over the same interval as f(x) since it consists of a polynomial and the cosine function.
To apply the MVT, you need to show that f'(x) satisfies two conditions:
a) f'(x) is continuous over the interval.
b) f'(x) equals zero at least once in the interval.
The first condition is already satisfied since f'(x) is continuous.
To satisfy the second condition, you need to find an interval for which f'(x) equals zero. Solve for x in the equation f'(x) = 0:
5 - 3cos(3x) = 0.
3cos(3x) = 5.
cos(3x) = 5/3.
The equation cos(3x) = 5/3 has a solution if and only if the right-hand side value of 5/3 lies within the range [-1, 1]. Since 5/3 is greater than 1, the equation has no real solutions. Consequently, f'(x) = 0 has no real roots, implying that f(x) = 5x - 7 - sin(3x) has no critical points.
Since we know the function is continuous and has at least one root, and it has no critical points to create additional roots, we can conclude that it has exactly one real zero in the interval where we established its existence previously.