Question

For the function y = (x² + 4)(x³ - 9x) at (-3, 0):

(a) The slope of the tangent line at (-3, 0) =
(b) The instantaneous rate of change of the function at (-3, 0) =

Answer 1

The slope of the tangent line and the instantaneous rate of change of the function y = (x² + 4)(x³ - 9x) at the point (-3, 0) are both -9, obtained by finding the derivative and plugging in the value x=-3.

Step-by-step explanation:

The question asks for the slope of the tangent line and the instantaneous rate of change of the function y = (x² + 4)(x³ - 9x) at the point (-3, 0). Both of these can be found by first taking the derivative of the function to find the general slope of the tangent at any point on the curve.

To find the derivative, we can use the product rule: if ∗(uv) = u∗v + v∗u, for our function u=x²+4 and v=x³-9x. This leads to ∗y = (x²+4)(3x²-9) + (2x)(x³-9x). Simplifying the derivative, we find ∗y = 3x´ - 9x² + 8x³ - 18x. We can now plug in the value x=-3 to find the derivative at that point, which is ∗y(-3) = 3(-3)´ - 9(-3)² + 8(-3)³ - 18(-3).

After calculating, we find that the slope of the tangent line and the instantaneous rate of change at the point (-3, 0) is equal to -9, since all terms involving -3 to even powers cancel out. Therefore:

1. The slope of the tangent line at (-3, 0) is -9.
2. The instantaneous rate of change of the function at (-3, 0) is -9.

Note that the slope of the tangent line and the instantaneous rate of change are the same value at a given point on the function.