Consider the following function. f(x) = ln(1 + 2x), a = 2, n = 3, 1.8 ≤ x ≤ 2.2 (a) Approximate f by a Taylor polynomial with de
gree n at the number a. T3(x) = (b) Use Taylor's Inequality to estimate the accuracy of the approximation f(x) ≈ Tn(x) when x lies in the given interval. (Round your answer to six decimal places.) |R3(x)| ≤
1 answer:
You might be interested in
We are given a relationship between the sides of a rectangle, that is, the length of one of its sides is 5 less two times its width, and we are asked to find an expression for the area. Let's remember that the area of a rectangle is equal to the product of the length of its side by its width. Let "w" be the length of the rectangle and "L" its lenght, then the area is given by the following formula:
We can use the relationship given in the problem, that is, its length being five less two times its width, that is:
Replacing in the formula for the area we get:
Now we use the distributive law:
1) Substituting into point-slope form, the equation of the line is y-6=⅓(x-3), which rearranges to:
- y-6 = ⅓x - 1
- y = ⅓x + 5
So, we can now substitute in the coordinates of each of the options to see which point lies on the line.
- 3 = ⅓(6) + 5 -> 3 = 7, which is false.
- 6 = ⅓(7) + 5 -> 6 = 22/3, which is false.
- -3 = ⅓(-3) + 5 -> -3 = 4, which is false.
- 3 = ⅓(-6) + 5 -> 3 = 3, which is true.
So, the answer is (4) (-6, 3)
2) Substituting into point-slope form, the equation of the line is y - 5 = ¾(x-2), which rearranges to:
- y - 5 = 0.75x - 1.5
- y = 0.75x + 3.5
So, we can now substitute in the coordinates of each of the options to see which point lies on the line.
- 8 = 0.75(6)+3.5 -> 8 = 8, which is true.
- 9 = 0.75(5) + 3.5 -> 9 = 7.25, which is false.
- 1 = 0.75(-1) + 3.5 -> 1 = 2.75, which is false.
- 2 = 0.75(6) + 3.5 -> 2 = 8, which is false.
So, the answer is (1) (6, 8).
