Question

show how to use the intermediate value theorem to show that the equation (x^2 - sqrt{2x} = 1) has a solution between 0 and 2.

let (f(x) = x^2 - sqrt{2x}). in order for the intermediate value theorem to apply we must first check that (f) is select an answer on the interval (0,2). you should verify this and be able to explain why it is the case.

we check the value of (f) at the left endpoint of the interval (0,2):
(f(\\) () = ) .

and we check the value of (f) at the right endpoint of the interval (0,2):

Explanation:

Step1: Identify the required property for the theorem

continuous

Step2: Evaluate the function at the left endpoint

$$f(0) = 0^2 - \sqrt{2(0)} = 0$$

Step3: Evaluate the function at the right endpoint

$$f(2) = 2^2 - \sqrt{2(2)} = 4 - 2 = 2$$

Answer:

continuous
\(f(0) = 0\)
\(f(2) = 2\)