Question

How do you use the first and second derivatives to sketch `y=2x^3- 3x^2 -180x`?

Answer 1

`dy/dx(2x^3-3x^2-180x) = 6[x^2 - x -30]`

`(d^2y)/dx^2(2x^3-3x^2-180x) = 12x-6`

Graph is also available with relevant details.

Explanation:

We are given the function

`color(red)(y = 2x^3-3x^2-180x)`

`color(green)(Step.1)`

We will find the First Derivative and set it equal to ZERO.

`rArr dy/dx(2x^3-3x^2-180x)`

`rArr 2*dy/dx(x^3)-3*dy/dx(x^2)-180*dy/dx(x)`

`rArr 2*3x^2-3*2x^1-180*1`

`rArr 6x^2-6x - 180`

We will factor out the GCF

`rArr 6*[ x^2-x - 30 ]`

This is the First Derivative of `color(red)(y = 2x^3-3x^2-180x)`

Hence,

`color(blue)((dy)/dx (2x^3-3x^2-180x) = 6*[ x^2-x - 30 ]`

We will now set this equal to Zero.

`:. 6*[ x^2-x - 30 ]= 0`

`rArr [ x^2-x - 30 ] = 0`

We will split the x-term to get

`rArr [ x^2+5x-6x - 30 ] = 0`

`rArr x*(x+5) - 6 *(x+5) = 0`

`rArr (x+5) *(x-6) = 0`

Roots or Zeros are found at

`color(blue)(x = -5 or x = +6)`

We can now say that ...

the Critical Numbers in this graph are ( - 5 and 6 )

`color(green)(Step.2)`

Let us now place these critical values on a Number Line and then generate a Sign Chart

Sign Chart for our First Derivative is given below:

We observe the following for the First Derivative Test:

1. If the first derivative is Positive [ f'(x) > 0 ] then our Original Function is Increasing.

2. If the first derivative is Negative [ f'(x) < 0 ] then our Original Function is Decreasing.

At . . `color(red)(x = +6)` our function has a maximum

At . . `color(red)(x = -5)` our function has a minimum

`color(green)(Step.3)`

We will now find the Second Derivative

We have

`color(blue)((dy)/dx (2x^3-3x^2-180x) = 6*[ x^2-x - 30 ]`

We must now find

`color(blue)((d^2y)/dx^2 (2x^3-3x^2-180x)`

`color(blue)((dy)/dx [6*( x^2-x - 30 )]`

That is we must differentiate our first derivative

`rArr 6*[d/dx (x^2) +d/dx (-x) + d/dx (-30) ]`

`rArr 6*[(2x) -1 + 0]`

`rArr 6*[2x -1]`

This is our Second Derivative

We will set the Second Derivative Equal to Zero

`rArr 6*[2x -1] = 0`

`rArr [2x -1]= 0`

`rArr 2x = 1`

`rArr x = 1/2`

`color(blue)((d^2y)/dx^2 (2x^3-3x^2-180x) = 6*[2x -1]`

Critical Point for the second derivative is `1/2`

`color(green)(Step.4)`

Let us now place these values on a Number Line and then generate a Sign Chart

Sign Chart for our Second Derivative is given below:

We observe the following for the Second Derivative Test:

1. If the second derivative is Positive [ f'(x) > 0 ] then our Original Function is Concave Up.

2. If the first derivative is Negative [ f'(x) < 0 ] then our Original Function is Concave Down.

`color(green)(Step.5)`

The graph below is for the Original Function

`color(red)(y = 2x^3-3x^2-180x)`

Study the graph and compare the results obtained from the first derivative and the second derivative tests.

Hope this helps.