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

Dec 30, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 {x}^{3} - 3 {x}^{2} - 180 x\right) = 6 \left[{x}^{2} - x - 30\right]$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 \left(2 {x}^{3} - 3 {x}^{2} - 180 x\right) = 12 x - 6$

Graph is also available with relevant details.

#### Explanation:

We are given the function

$\textcolor{red}{y = 2 {x}^{3} - 3 {x}^{2} - 180 x}$

$\textcolor{g r e e n}{S t e p .1}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 {x}^{3} - 3 {x}^{2} - 180 x\right)$

$\Rightarrow 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{3}\right) - 3 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2}\right) - 180 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \left(x\right)$

$\Rightarrow 2 \cdot 3 {x}^{2} - 3 \cdot 2 {x}^{1} - 180 \cdot 1$

$\Rightarrow 6 {x}^{2} - 6 x - 180$

We will factor out the GCF

$\Rightarrow 6 \cdot \left[{x}^{2} - x - 30\right]$

This is the First Derivative of $\textcolor{red}{y = 2 {x}^{3} - 3 {x}^{2} - 180 x}$

Hence,

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

We will now set this equal to Zero.

$\therefore 6 \cdot \left[{x}^{2} - x - 30\right] = 0$

$\Rightarrow \left[{x}^{2} - x - 30\right] = 0$

We will split the x-term to get

$\Rightarrow \left[{x}^{2} + 5 x - 6 x - 30\right] = 0$

$\Rightarrow x \cdot \left(x + 5\right) - 6 \cdot \left(x + 5\right) = 0$

$\Rightarrow \left(x + 5\right) \cdot \left(x - 6\right) = 0$

Roots or Zeros are found at

$\textcolor{b l u e}{x = - 5 \mathmr{and} x = + 6}$

We can now say that ...

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

$\textcolor{g r e e n}{S t e p .2}$

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 . . $\textcolor{red}{x = + 6}$ our function has a maximum

At . . $\textcolor{red}{x = - 5}$ our function has a minimum

$\textcolor{g r e e n}{S t e p .3}$

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

$\Rightarrow 6 \cdot \left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(- x\right) + \frac{d}{\mathrm{dx}} \left(- 30\right)\right]$

$\Rightarrow 6 \cdot \left[\left(2 x\right) - 1 + 0\right]$

$\Rightarrow 6 \cdot \left[2 x - 1\right]$

This is our Second Derivative

We will set the Second Derivative Equal to Zero

$\Rightarrow 6 \cdot \left[2 x - 1\right] = 0$

$\Rightarrow \left[2 x - 1\right] = 0$

$\Rightarrow 2 x = 1$

$\Rightarrow x = \frac{1}{2}$

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

Critical Point for the second derivative is $\frac{1}{2}$

$\textcolor{g r e e n}{S t e p .4}$

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.

$\textcolor{g r e e n}{S t e p .5}$

The graph below is for the Original Function

$\textcolor{red}{y = 2 {x}^{3} - 3 {x}^{2} - 180 x}$

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

Hope this helps.