How do you find the critical points to sketch the graph #h(x)=27x-x^3#?

Answer 1

Critical Points are:#color(blue)[(x = +3), (x = -3) #

Graph of #color(blue)(h(x) = 27x - x^3)# is also available with this solution.

Given:

#color(red)(h(x) = 27x - x^3)#

We need to find the Critical Points

Definition of Critical Points:

Let #color(blue)f# be a function and let #color(blue)c# be a point in its domain.

We call #color(blue)c# a Critical Point if

(1) #f'(c) = 0# or

(2) #f'(c)# is undefined.

Also not that,

(3) the derivative gives us the slope of the tangent line

(4) the Critical Points are points where the slope of the tangent line is ZERO

#color(green)(Step.1)#

Given:

#color(red)(h(x) = 27x - x^3)#

We will find the derivative first.

We are differentiating a polynomial. Hence, it is easy.

Note that #d/(dx)(x^n) = n*x^(n-1)#

#h'(x) = 27 - 3x^2#

#color(green)(Step.2)#

We will set this derivative equal to ZERO, to find our Critical Points.

#h'(x) = 27 - 3x^2 = 0#

Subtract #27# from both sides.

#rArr 27-3x^2 - 27 = 0 - 27#

#rArr cancel 27-3x^2 - cancel 27 = 0 - 27#

#rArr -3x^2 = - 27#

Divide both sides by #color(red)(-1)#

#rArr (-3x^2)/color(red)(-1) = (- 27)/color(red)(-1)#

#rArr 3x^2 = 27#

Divide both sides by 3

#rArr (3x^2)/3 = 27/3#

#rArr (cancel 3x^2)/cancel 3 = 27/3#

#rArr x^2 = 9#

Taking Square Root on both sides

#sqrt((x^2)) = +- sqrt(9)#

#rArr x = +- sqrt(9)#

#rArr x = +- 3#

Hence our Critical Points are #color(blue)[(x = +3), (x = -3) #

#color(green)(Step.3)#

Please analyze the graph below:

Please observe the following on the graph:

At the Critical Points, tangent lines are horizontal in this example

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Answer 2

To find the critical points of the function ( h(x) = 27x - x^3 ) in order to sketch its graph, follow these steps:

  1. Compute the derivative of ( h(x) ) to find critical points: [ h'(x) = \frac{d}{dx}(27x - x^3) ] [ h'(x) = 27 - 3x^2 ]

  2. Set the derivative equal to zero and solve for ( x ) to find critical points: [ 27 - 3x^2 = 0 ] [ 3x^2 = 27 ] [ x^2 = 9 ] [ x = \pm 3 ]

  3. Determine the nature of critical points by analyzing the sign of the derivative around each critical point.

    • For ( x = -3 ):

      • Substitute ( x = -3 ) into ( h'(x) ): ( h'(-3) = 27 - 3(-3)^2 = 27 - 27 = 0 ).
      • Since the derivative changes sign from negative to positive at ( x = -3 ), it indicates a local minimum.
    • For ( x = 3 ):

      • Substitute ( x = 3 ) into ( h'(x) ): ( h'(3) = 27 - 3(3)^2 = 27 - 27 = 0 ).
      • Since the derivative changes sign from positive to negative at ( x = 3 ), it indicates a local maximum.
  4. Plot the critical points ( (-3, h(-3)) ) and ( (3, h(3)) ) on the graph.

  5. Determine the behavior of the function as ( x ) approaches positive and negative infinity:

    • As ( x ) approaches positive infinity, ( h(x) ) approaches negative infinity because of the term ( -x^3 ).
    • As ( x ) approaches negative infinity, ( h(x) ) approaches positive infinity because the term ( 27x ) dominates.
  6. Sketch the graph based on the behavior around critical points and at infinity.

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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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