Give an example of an increasing function whose domain is the interval [0,1]but whose range does not equal the interval[f(0),f(1)]

please show me the work

well, f(x) = 1/(1-x)

has range = [1,oo) but it's sort of a bad candidate because f(1) is not defined, so [f(0),f(1)] does not exist.

Gotta think more on this one.

To find an example of an increasing function with the domain [0,1], but with a range that does not equal the closed interval [f(0), f(1)], we can follow these steps:

Step 1: Define a function with the desired domain:
Let's define a function f(x) = x^2, where x belongs to the interval [0,1].

Step 2: Calculate the range of the function:
To find the range of f(x), we need to find the minimum and maximum values that f(x) can take on the interval [0,1].

At x = 0, f(0) = (0)^2 = 0.
At x = 1, f(1) = (1)^2 = 1.

Thus, the range of f(x) is [0,1].

Step 3: Demonstrate that the range is not equal to [f(0), f(1)]:
We have found that the range of f(x) is [0,1]. However, we need to show that the range is not equal to the interval [f(0), f(1)].

f(0) = 0^2 = 0, and f(1) = 1^2 = 1. So, the interval [f(0), f(1)] is [0, 1].

Since the range [0,1] is equal to the interval [f(0), f(1)], we need to choose a different function to obtain a range that is not equal to [f(0), f(1)].

Example:
Let's define a new function g(x) = x^2 - x, where x belongs to the interval [0,1].

Step 4: Calculate the range of the new function:
Similar to Step 2, we find the minimum and maximum values of g(x) on the interval [0,1].

At x = 0, g(0) = (0)^2 - 0 = 0.
At x = 1, g(1) = (1)^2 - 1 = 0.

Hence, the range of g(x) is {0}.

Step 5: Show that the range is not equal to [f(0), f(1)]:
The range is {0}, but the interval [f(0), f(1)] is [0, 1].

Since {0} is not equal to [0, 1], we have found an example of an increasing function f(x) = x^2 - x, defined on the interval [0,1], where the range is not equal to [f(0), f(1)].