Final answer:
To prove that a function increases throughout an interval when it is positive, we can use the mean value theorem. The derivative of a function can also be used to determine the intervals on which the function increases or decreases. By finding the x-intercepts and y-intercept, along with information about the increasing and decreasing intervals, we can sketch the graph of the function.
Step-by-step explanation:
To prove that f increases throughout (a, b) when f(x) > 0 for every x in (a, b), we can use the mean value theorem. Since f is continuous on [a, b] and differentiable on (a, b), the mean value theorem states that there exists at least one point c in (a, b) where the derivative of f(x) is equal to the secant slope between a and b. Since f(x) > 0 for every x in (a, b), the derivative of f(x) must be positive throughout (a, b), which means f is increasing throughout (a, b).
To find the intervals on which f(x) = 2 + x - x^2 increases and decreases, we can take the derivative of f(x). The derivative of f(x) is equal to 1 - 2x. Setting this derivative equal to 0 and solving for x, we find x = 1/2. Thus, f(x) decreases when x < 1/2 and increases when x > 1/2. Therefore, f(x) = 2 + x - x^2 decreases on the interval (-∞, 1/2) and increases on the interval (1/2, +∞).
To sketch the graph of f(x) = 2 + x - x^2, we first find the x-intercepts by setting f(x) equal to 0 and solving for x. Since f(x) = 2 + x - x^2, we have 2 + x - x^2 = 0. This quadratic equation can be factored as (x - 2)(x + 1) = 0. Therefore, the x-intercepts are x = 2 and x = -1.
We can also find the y-intercept by setting x equal to 0 in f(x). When x = 0, f(x) = 2 + 0 - 0^2 = 2. Therefore, the y-intercept is (0, 2).
Using these points and the information about the increasing and decreasing intervals, we can sketch the graph of f(x) = 2 + x - x^2.