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

Answer:

(a) $-65$

(b) $-65$

Step-by-step 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.

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

$y'=\frac{d}{dx}(x^3-3)\cdot (x^2-5x+1)+(x^3-3)\cdot \frac{d}{dx}(x^2-5x+1)$

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

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

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

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 $(2,-25)$.