Use intermediate value theorem to show there is a root to 2x^3 + x^2 - 2 = 0 on [0,1]
f(0) = -2
f(1) = 2+1-2 = 1
mmmhhh, how did the graph get from -2 to +1 ?
To apply the Intermediate Value Theorem, we need to show that the function 2x^3 + x^2 - 2 is continuous on the interval [0,1] and takes on both positive and negative values.
To begin, let's check the continuity of the function on [0,1]. The polynomial function 2x^3 + x^2 - 2 is continuous since it is a combination of power functions (x raised to a constant power) and constant functions.
Next, we'll evaluate the function at the endpoints of the interval [0,1].
When x = 0, we have:
2(0)^3 + (0)^2 - 2 = -2.
When x = 1, we have:
2(1)^3 + (1)^2 - 2 = 1.
Since the function takes on a negative value at x = 0 and a positive value at x = 1, we can conclude that it must pass through the x-axis at some point in the interval [0,1], according to the Intermediate Value Theorem. Therefore, there is at least one root of the equation 2x^3 + x^2 - 2 = 0 between x = 0 and x = 1.