Math Problem Statement

What are the coordinates of a relative minimum for y = -x^4 - x^3 + 8x^2 + 6x - 5 to 2 decimal places?

Solution

To find the coordinates of the relative minimum for the function y=−x4−x3+8x2+6x−5y = -x^4 - x^3 + 8x^2 + 6x - 5, we will follow these steps:

1. Differentiate the function

We need to find the first derivative y′y' to locate the critical points (where y′=0y' = 0) and the second derivative y′′y'' to determine if those points are minima.

Given: y=−x4−x3+8x2+6x−5y = -x^4 - x^3 + 8x^2 + 6x - 5

The first derivative is:

= -4x^3 - 3x^2 + 16x + 6$$ ### 2. Solve $$y' = 0$$ to find critical points We solve $$-4x^3 - 3x^2 + 16x + 6 = 0$$ for $$x$$. This is a cubic equation, so we'll use a numerical method (e.g., the Newton-Raphson method or a graphing calculator) to find approximate solutions. ### 3. Use the second derivative to confirm the relative minimum After finding the critical points from step 2, we compute the second derivative $$y''$$ to apply the second derivative test. If $$y''(x) > 0$$, it's a local minimum. Once the critical points are determined and classified as minima, we can find the corresponding $$y$$-values by plugging the $$x$$-values back into the original function to get the full coordinates of the relative minimum. Let me solve this for you: Solving $$y' = 0$$: The solutions are approximately $$x = -0.36$$, $$x = 1.85$$, and $$x = -1.29$$. Now, using the second derivative test, it turns out that $$x = -0.36$$ corresponds to a relative minimum. ### 4. Compute the corresponding $$y$$-value Substitute $$x = -0.36$$ into the original function: $$y(-0.36) = -(-0.36)^4 - (-0.36)^3 + 8(-0.36)^2 + 6(-0.36) - 5 \approx -6.09$$ Thus, the coordinates of the relative minimum are $$(-0.36, -6.09)$$. ### Answer: The correct choice is **(-0.36, -6.09)**. Would you like a deeper breakdown of any steps? Here are five follow-up questions: 1. How do we interpret a second derivative test for minima and maxima? 2. Can the cubic function $$y' = 0$$ have more than three real roots? 3. How do we apply the Newton-Raphson method to solve higher-degree polynomials? 4. How do relative minima differ from absolute minima? 5. Can the sign of the leading term in a polynomial tell us about its end behavior? **Tip:** When solving cubic or higher-degree polynomials, numerical or graphical methods are often more efficient than attempting to solve them algebraically.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Second Derivative Test

Formulas

First derivative: y' = -4x^3 - 3x^2 + 16x + 6
Second derivative: y'' = -12x^2 - 6x + 16

Theorems

Second Derivative Test

Suitable Grade Level

Grades 11-12 or Calculus 1

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