Mathematics: 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

Q =11All you have to do is plug in 11 for Q.11-3 / 2 = 8/2 = 4...

For the function y = (x^3 − 3)(x^2 − 5x + 1) at

Ответ:

mayfieldashley2437

13.01.2022

Mathematics

(a)

(b)

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 , therefore, the instantaneous rate of change of the function is also at point (2,-25) .

0,0(0 оценок)

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

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question