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Chapter 4: Problem 23

a. Write the derivative formula. b. Locate any relative extreme points and identify the extreme as a maximum or minimum. $$ h(x)=x^{3}-8 x^{2}-6 x $$

Short Answer

Expert verified
The relative maximum is at \( x \approx -0.35 \) and the relative minimum is at \( x \approx 5.69 \).

Step by step solution

01

Find the Derivative

To find the derivative of the given function \( h(x) = x^3 - 8x^2 - 6x \), apply the power rule of differentiation. The derivative \( h'(x) \) is given by: \( h'(x) = 3x^2 - 16x - 6 \).
02

Set the Derivative to Zero

To find the critical points, set the derivative \( h'(x) = 3x^2 - 16x - 6 \) equal to zero. Solve \( 3x^2 - 16x - 6 = 0 \) to find the potential extreme points.
03

Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = -16 \), and \( c = -6 \). Calculate the discriminant: \( b^2 - 4ac = 256 + 72 = 328 \). Then calculate \( x = \frac{-(-16) \pm \sqrt{328}}{6} \), giving solutions \( x_1 \approx 5.69 \) and \( x_2 \approx -0.35 \).
04

Determine the Nature of Each Critical Point

Use the second derivative test to determine if each critical point is a maximum or minimum. First, find the second derivative: \( h''(x) = 6x - 16 \). Evaluate \( h''(5.69) \) and \( h''(-0.35) \). Since \( h''(5.69) > 0 \), \( x = 5.69 \) is a local minimum. Since \( h''(-0.35) < 0 \), \( x = -0.35 \) is a local maximum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental tool in calculus for differentiating expressions of the form \( x^n \). It states that if you have a function \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \). This rule makes it straightforward to find the derivatives of polynomial functions, such as the one in our original exercise: \( h(x) = x^3 - 8x^2 - 6x \).
To apply the power rule, individually differentiate each term in the function:
  • For \( x^3 \), the derivative is \( 3x^2 \) (using \( n=3 \)).
  • For \( -8x^2 \), the derivative is \( -16x \) (using \( n=2 \)).
  • For \( -6x \), the derivative is \( -6 \) (using \( n=1 \)).
Therefore, the derivative \( h'(x) \) is \( 3x^2 - 16x - 6 \), which is essential for finding critical points and analyzing function behavior.
Critical Points
Critical points are where the function's derivative is zero or undefined. These points are essential in determining where a function may have relative extrema (local maximums or minimums).
In the exercise, the function \( h(x) = x^3 - 8x^2 - 6x \) has the derivative \( h'(x) = 3x^2 - 16x - 6 \). To find critical points, we solve the equation \( h'(x) = 0 \).
This involves solving the quadratic equation:
  • \( 3x^2 - 16x - 6 = 0 \)
Solving this gives the potential values for \( x \) where the function has a turning point. These turning points are where the function changes from increasing to decreasing or vice versa, indicating possible locations for maxima or minima.
Quadratic Formula
The quadratic formula is used to find the roots of a quadratic equation \( ax^2 + bx + c = 0 \). It is one of the most reliable methods to solve quadratic equations and is given by the formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our exercise, the quadratic \( 3x^2 - 16x - 6 = 0 \) can be solved using this formula. Here, \( a = 3 \), \( b = -16 \), and \( c = -6 \).
Calculate:
  • Discriminant: \( b^2 - 4ac = 256 + 72 = 328 \).
  • Substitute values into the quadratic formula to find \( x_1 \approx 5.69 \) and \( x_2 \approx -0.35 \).
These solutions are the \( x \)-coordinates of the possible extrema of the function.
Second Derivative Test
Once we have identified the critical points, the second derivative test helps determine the nature of these points. This test examines the concavity of the function at its critical points.
The second derivative \( h''(x) \) of our function \( h(x) \) is given by differentiating \( h'(x) = 3x^2 - 16x - 6 \) again, resulting in \( h''(x) = 6x - 16 \).
For each critical point \( x_1 \approx 5.69 \) and \( x_2 \approx -0.35 \), we evaluate the second derivative:
  • If \( h''(x) > 0 \), the function is concave up, indicating a local minimum.
  • If \( h''(x) < 0 \), the function is concave down, indicating a local maximum.
Evaluating:
  • At \( x \approx 5.69 \), \( h''(x) > 0 \), so it is a local minimum.
  • At \( x \approx -0.35 \), \( h''(x) < 0 \), so it is a local maximum.
This method can help confirm the nature of critical points efficiently.

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