At the indicated point for the function, find the following. A graphing utility's numerical derivative feature can be used to check your work.
y = (x3 + 2x)3 at x = 3
(a) Find the slope of the tangent line.
(b) Find the instantaneous rate of change of the function.
To find the slope of the tangent line and the instantaneous rate of change of the function at a specific point, you can use the concept of the derivative.
(a) The slope of the tangent line at a point is the same as the derivative of the function at that point. To find the derivative of the function, you can take the derivative with respect to x.
To find the derivative, you can use the power rule, which states that the derivative of x^n is n*x^(n-1).
Taking the derivative of the function y = (x^3 + 2x)^3 with respect to x:
dy/dx = 3 * (x^3 + 2x)^2 * (3x^2 + 2)
Now, plug in the given x-coordinate, x = 3, into the derivative expression:
dy/dx = 3 * (3^3 + 2*3)^2 * (3*3^2 + 2)
dy/dx = 3 * (27 + 6)^2 * (27 + 2)
dy/dx = 3 * (33)^2 * (29)
dy/dx = 3 * 1089 * 29
dy/dx = 9801
Therefore, the slope of the tangent line at x = 3 is 9801.
(b) The instantaneous rate of change of the function at a point is also the same as the derivative of the function at that point. So, the instantaneous rate of change is also 9801.