The function f is continuous on the closed interval [0,2] and has values that are given in the table below
x = 0| 1 | 2
____________
f(x) = 1| k | 2
The equation f(x) = 1/2 must have at least two solutions in the interval [0,2] if K = ?
The answer is 0, but there is also a choice option of 1/2. I don't understand how to get the answer.
Thanks!
Oh boy, math problems! Let's dive in and see if I can add some humor to make it more fun!
So, we want to find the value of k in order for the equation f(x) = 1/2 to have at least two solutions in the interval [0,2].
Looking at the table, we see that f(x) is 1 at x = 0 and 2 at x = 2. But we don't know the value at x = 1, which is represented by the mysterious k.
Now, if we draw a graph of f(x), it will be a continuous curve between x = 0 and x = 2. And we're looking for the value of k that will make the graph intersect the horizontal line y = 1/2 at least twice.
Imagine the graph as a roller coaster ride. The line y = 1/2 is like a safety bar that we want the roller coaster to go under at least twice. We don't want any flying clowns here!
Now, in order for the roller coaster to cross the safety bar twice, it needs to go above the safety bar first and then come back down underneath it. Think of it as the roller coaster doing some crazy loop-de-loops!
If we look at the table again, we can see that the value of f(x) at x = 1 is represented by k. If k is less than 1/2, the graph will go above the safety bar and come back down twice, giving us the two solutions we need.
So, the value of k has to be less than 1/2 in order for the equation f(x) = 1/2 to have at least two solutions.
Therefore, the answer is indeed 0, not 1/2. The roller coaster is doing some serious acrobatics to make this happen!
To find the value of k that makes the equation f(x) = 1/2 have at least two solutions in the interval [0,2], we need to analyze the behavior of the function on that interval based on the given table.
First, let's consider the given values of f(x) in the table. At x = 0, f(x) = 1, and at x = 2, f(x) = 2. We also know that f(x) is continuous on the closed interval [0, 2].
Since f(x) is continuous on [0, 2], and the function values at the endpoints of the interval are different (f(0) = 1 and f(2) = 2), by the Intermediate Value Theorem, there must exist at least one value c in the open interval (0, 2) such that f(c) = 1/2. This means that the equation f(x) = 1/2 has at least one solution in the interval (0, 2).
But we want to find the value of k that makes the equation have at least two solutions in the interval [0, 2]. For that to happen, the function f(x) must have at least one more value between 0 and 2 where f(x) = 1/2. Since the function values at the endpoints are different, f(x) must pass through 1/2 somewhere between 0 and 2.
From the given table, we see that f(x) has the value k at x = 1. So, for f(x) to have at least two solutions, k must be equal to 1/2. This is because when k = 1/2, the function f(x) will have a value of 1/2 at x = 1 as well (which is between 0 and 2), giving us the second solution of the equation f(x) = 1/2 in the interval [0, 2].
Therefore, the value of k that makes the equation f(x) = 1/2 have at least two solutions in the interval [0, 2] is k = 1/2.
To find the value of k that satisfies the equation f(x) = 1/2 in the interval [0,2], we need to analyze the given table.
Given that the function f is continuous on the closed interval [0,2], we can determine its behavior by examining the values of f(x) at the interval endpoints and in between.
In the table, we see that f(0) = 1 and f(2) = 2.
To satisfy the equation f(x) = 1/2, the function f(x) would need to have at least two x-values in the interval [0,2] where f(x) is equal to 1/2.
If we choose k to be 0, then f(1) = 0. However, f(0) = 1, and f(2) = 2, which means that f(x) = 1/2 does not have at least two solutions in the interval [0,2].
On the other hand, if we choose k to be 1/2, then we have f(1) = 1/2. In this case, f(0) = 1, f(1) = 1/2, and f(2) = 2. Thus, f(x) = 1/2 has at least two solutions in the interval [0,2] when k = 1/2.
Therefore, the answer is k = 1/2, not 0 as stated.
For f(x)=1/2 to have two solutions on the interval [0,2], a horizontal line through y=1/2 must intersect f(x) two times or more on the interval.
This will happen if k<=1/2. In the limit where k=1/2, the horizontal line will be tangent to f(x), and is considered to have two solutions.