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.

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.

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Answer 2

To sketch the graph of ( y = 2x^3 - 3x^2 - 180x ), we can use the first and second derivatives to analyze the behavior of the function.

  1. First, find the first derivative ( y' ) by differentiating ( y ) with respect to ( x ).
  2. Next, find the critical points of the function by setting the first derivative equal to zero and solving for ( x ).
  3. Then, use the second derivative test to determine the nature of these critical points (whether they correspond to local maxima, local minima, or points of inflection).
  4. Finally, plot the graph using this information.

Here are the steps in more detail:

  1. First derivative: [ y' = \frac{dy}{dx} = 6x^2 - 6x - 180 ]

  2. Critical points: Setting ( y' = 0 ), we solve for ( x ): [ 6x^2 - 6x - 180 = 0 ] [ x^2 - x - 30 = 0 ] [ (x - 6)(x + 5) = 0 ]

So, the critical points are ( x = 6 ) and ( x = -5 ).

  1. Second derivative: [ y'' = \frac{d^2y}{dx^2} = 12x - 6 ]

  2. Evaluate the second derivative at the critical points: [ y''(6) = 12(6) - 6 = 66 ] [ y''(-5) = 12(-5) - 6 = -66 ]

Since ( y''(6) > 0 ), the function has a local minimum at ( x = 6 ). Since ( y''(-5) < 0 ), the function has a local maximum at ( x = -5 ).

  1. Plot the points and sketch the graph:
  • At ( x = -5 ), there is a local maximum.
  • At ( x = 6 ), there is a local minimum.
  • As ( x ) approaches negative or positive infinity, the function increases without bound.

With this information, you can sketch the graph of ( y = 2x^3 - 3x^2 - 180x ), noting the local maximum and minimum points and the overall behavior of the function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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