54.9k views
5 votes
Applications of Limits, showing each formula you used, and all of your work.

Applications of Limits, showing each formula you used, and all of your work.-example-1
User Secretlm
by
9.1k points

1 Answer

2 votes

Answer:

Explanation:

A. To find the average rate of change over the interval [1, 4], we need to calculate the slope of the secant line between two points on the curve.

The function is f(x) = x^3 - x + 2.

Let's calculate the end point value of the interval.

f(1) = (1)^3 - 1 + 2 = 2

f(4) = (4)^3 - 4 + 2 = 62

The average rate of change is obtained from the slope of the secant line.

Average rate of change = (f(4) - f(1)) / (4 - 1) = (62 - 2) / (4 - 1) = 60 / 3 = 20.

So the average rate of change over the interval [1, 4] is 20. B. To find the instantaneous rate of change at x=2, we need to take the derivative of the function and evaluate it at x=2.

The derivative of f(x) = x^3 - x + 2 is given by

f'(x) = 3x^2 - 1.

Let's evaluate f'(x) at x = 2.

f'(2) = 3(2)^2 - 1 = 3(4) - 1 = 12 - 1 = 11.

So the instantaneous rate of change at x = 2 is 11. C. To find the equation of the tangent line at x = 2, we need the slope of the tangent line and a point on the line.

The slope of the tangent line at x = 2 has already been calculated, which is 11.

To find the point on the line, evaluate the original function at x = 2.

f(2) = (2)^3 - 2 + 2 = 8 - 2 + 2 = 8.

So the point on the tangent line is (2, 8). Now we can write the tangent equation using the point and slope form of the line.

y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope.

Insert value:

y - 8 = 11(x - 2).

Simplify:

y - 8 = 11x - 22.

The tangent equation at x = 2 is

y = 11x - 14.

User Mondano
by
8.7k points

No related questions found