function intervals

We can take the value of f at a (i.e., f(a)) as first estimate for average value of the function. Divide [a, b] into two equal parts such that then the second estimate of the average value of the function can be taken as second estimate of the average value of the functionh..

Find the range of the function in interval notation. y = - x 2 + 2 => (- ∞, 2] or [- 2, ∞) or (- ∞, 2) or [2, ∞) or (- ∞, - 2]..

Use Newton's method to find the zero of the function in the interval [- 1, 0] to the nearest four decimal places. f ( x ) = ln (2 + x ) + x => - 0.4428 or - 0.4621 or - 0.3 or - 1.386 or - 0.5..

Find the interval on which the function y = ( x - 5) 2 is increasing. => [5, ∞) or [5, ∞] or [- 5, 5) or (5, ∞]..

Choose the interval on which the function y = | x + 2| + | x - 1| - 2 is decreasing. => [- 2, 2] or [- 2, ∞) or (- ∞, ∞) or (-∞, - 2]..

In which interval does the function f ( x ) = 2 x 3 - 3 x 2 - 12 x + 10 increase? => R - {0, 1, 2} or (- 1, 2) or (- ∞, - 1) ∪ (2, ∞) or (- ∞, - 1] ∪ [2, ∞) or (0, ∞)..

Find the value in the interval, which satisfies the Mean Value Theorem for the function f ( x ) = 1 - x on [- 8, 1]. => 0 or - 5 4 or 1 or - 4 5 or 3..

Find the value in the interval, which satisfies the Mean Value Theorem for the function f ( x ) = 1 + 1 x on [1, 4]. => 3 9 6 or 3 5 1 0 or - 2 or 2 or 0..

Find the value in the interval, which satisfies the Mean Value Theorem for the function f ( x ) = x - x 3 on [- 2, 1]. => 1 or - 1 or 0 or 6 or 1 3..

Find the value in the interval, which satisfies the Mean Value Theorem for the function f ( x ) = ( x - 1)( x - 2) on [0, 4]. => 3 or 1 or 5 2 or - 2 or 2..