Sorry...

Consider the function f(x) = 8.5 x − cos(x) + 2 on the interval 0 ¡Ü x ¡Ü 1. The Intermediate Value Theorem guarantees that there is a value c such that for which values of c and k? Fill in the following mathematical statements, giving an interval with non-zero length in each case.

For every k in ____ ¡Ü k ¡Ü___ ,
there is a c in 0 ¡Ü c ¡Ü 1
such that .

Sorry...

Again...
Before I posted the question, I could see the symbols...
Consider the function f(x) = 8.5 x − cos(x) + 2 on the interval 0 less than or equal to x less than or equal to 1. The Intermediate Value Theorem guarantees that there is a value c such that for which values of c and k? Fill in the following mathematical statements, giving an interval with non-zero length in each case.

For every k in ______ less than or equal to k less than or equal to ______ ,
there is a c in 0 less than or equal to c less than or equal to 1
such that .

"Consider the function f(x) = 8.5 x − cos(x) + 2 on the interval 0 less than or equal to x less than or equal to 1.

The Intermediate Value Theorem guarantees that there is a value c such that for 0≤c≤1 and f(0)≤k≤f(1), if f(1)≥f(0), and f(0)≥k≥f(1) if f(1)<f(0)."

In this case, f(0)<f(1), so the first case applies.

P.S.
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≤ & l e ;
≥ & g e ;

For every k in (-1.5, 10), there is a c in 0 ≤ c ≤ 1 such that .

To find the interval for the values of k guaranteed by the Intermediate Value Theorem for the given function, we need to analyze the behavior of the function over the specified interval.

First, let's determine the minimum and maximum values of the function f(x) = 8.5x - cos(x) + 2 over the interval 0 ≤ x ≤ 1.

To find the minimum and maximum values, we can take the derivative of the function and set it equal to zero to find critical points. Then we evaluate the function at these critical points and at the endpoints of the interval to determine the extreme values.

The derivative of f(x) with respect to x is: f'(x) = 8.5 + sin(x).

To find the critical points, we set f'(x) = 0:

8.5 + sin(x) = 0
sin(x) = -8.5

We notice that sin(x) can only be between -1 and 1, so we conclude that there are no solutions to this equation. Therefore, f(x) has no critical points in the interval [0, 1].

Now, we evaluate the function at the endpoints of the interval:

f(0) = 8.5(0) - cos(0) + 2 = 2
f(1) = 8.5(1) - cos(1) + 2 ≈ 10.45

So, the minimum value of f(x) in the interval is 2 and the maximum value is approximately 10.45.

Therefore, we can now fill in the statements:

For every k in [2, 10.45],
there is a c in 0 ≤ c ≤ 1
such that .

This means that for any value of k between 2 and approximately 10.45, there exists at least one value of c in the interval [0, 1] such that f(c) equals k, guaranteed by the Intermediate Value Theorem.