Question

For the function y = (x^3 − 3)(x^2 − 5x + 1) at (2, −25) find the following.

(a) the slope of the tangent line
(b) the instantaneous rate of change of the function

Answer 1

(a)
`-65`

(b)
`-65`

Explanation:

We have been a function
`y=(x^3-3)(x^2-5x+1)`. We are asked to find the instantaneous rate of change of the function and the slope of the tangent line at point
`(2,-25)`.

(a) First of all, we will find the derivative of our given function using product rule.

`(f\cdot g)'=f'\cdot g+f\cdot g'`

`y'=(d)/(dx)(x^3-3)\cdot (x^2-5x+1)+(x^3-3)\cdot (d)/(dx)(x^2-5x+1)`

`y'=3x^2\cdot (x^2-5x+1)+(x^3-3)\cdot 2x-5`

`y'=3x^4-15x^3+3x^2+2x^4-6x-5x^3+15`

`y'=5x^4-20x^3+3x^2-6x+15`

Now, we will substitute
`x=2` in our derivative function to find slope of tangent line as:

`y'=5(2)^4-20(2)^3+3(2)^2-6(2)+15`

`y'=5(16)-20(8)+3(4)-12+15`

`y'=80-160+12-12+15`

`y'=-65`

Therefore, the slope of the tangent line is -65 at point
`(2,-25)`.

(b) We know that instantaneous rate of change of the function at a point is equal to the derivative of the function at that point.

We already figured it out that derivative of our given function at
`x=2` is
`-65`, therefore, the instantaneous rate of change of the function is also
`-65` at point
`(2,-25)`.