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.