Suppose f(x)=-x^2+1, and the domain of the funtion is the real numbers. Give an example fo a number that is nto in the range of f.

0 and 1 are in the range. So is any real r less than 1

thanks but i have to find a number that is not in the range.

I did not know what your "nto" meant. Any number greater than 1 is NOT in the range

i was thinking that it would be any number less than O. why is that not correct?

Numbers less than zero ARE in the range of possible f(x) values. For example, if x = 2, f(x) = -3

To find an example of a number that is not in the range of f(x) = -x^2 + 1, we need to determine the range of the function.

The range of a function refers to the set of all possible outputs or values that the function can produce. In this case, we have a quadratic function with a negative coefficient in front of the x^2 term. This means that the graph of the function will be a downward-opening parabola.

To find the range, we can determine the vertex (maximum point) of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a = -1 and b = 0, so the x-coordinate of the vertex is x = -0 / 2(-1) = 0.

Substituting x = 0 into the function, we get f(0) = -(0)^2 + 1 = 1. Therefore, the vertex of the parabola is (0, 1).

Since the parabola opens downwards, the maximum value of the function occurs at the vertex, and the range of f(x) is all values less than or equal to 1.

To find an example of a number not in the range, we can choose any number greater than 1. For instance, the number 2 is not in the range of f(x) because f(x) can never be greater than 1.

So, 2 is an example of a number that is not in the range of f(x) = -x^2 + 1.