Final answer:
The range of values of c for which all values of the function f(x)=x² cx-1/2x²-3x 2 belong to the interval (-1, 2) is c ∈ (-∞, 0) U (0, ∞).
Step-by-step explanation:
To find all possible values of c for which all values of the function f(x)=x² cx-1/2x²-3x 2 belong to the interval (-1, 2), we need to find the range of values of c for which the function f(x) is always less than 2 and greater than -1.
We can rewrite the function as f(x) = (x² cx)/(2x²-6x) - 1.
We can simplify this expression to f(x) = (cx)/(2x-6) - 1.
The denominator of this expression is equal to 2(x-3), which means that the function is undefined at x=3.
Since the function is continuous on the interval (-1, 2), we can find the range of values of c for which the function is always less than 2 and greater than -1 by finding the maximum and minimum values of the function on this interval.
To find the maximum and minimum values of the function, we can take the derivative of the function and set it equal to zero. The derivative of the function is f’(x) = (2cx-6c-4x)/(2x-6)².
Setting this equal to zero, we get 2cx-6c-4x = 0, or c = (2x)/(3-x).
The critical points of the function are x=0 and x=3. We can test the intervals (-1, 0), (0, 3), and (3, 2) to determine the sign of f’(x) in each interval. We find that f’(x) is negative in the interval (-1, 0), positive (0, 3), and negative in the interval (3, 2).
Therefore, the maximum value of the function on the interval (-1, 2) occurs at x=0, and the minimum value of the function on the interval (-1, 2) occurs at x=3.
We can substitute these values of x into the function to find the corresponding maximum and minimum values of the function.
Substituting x=0, we get f(0) = 0 - 1 = -1. Substituting x=3, we get f(3) = 9c/0, which is undefined.
Therefore, the range of values of c for which all values of the function f(x)=x² cx-1/2x²-3x 2 belong to the interval (-1, 2) is c ∈ (-∞, 0) U (0, ∞).