Math Problem Statement

Find the zeros, maximums, and minimums of the polynomial: f(x) = 2x^5 + 3x^4 - 6x^2 + 4.

Solution

The given polynomial is:

f(x)=2x5+3x46x2+4f(x) = 2x^5 + 3x^4 - 6x^2 + 4

To solve the problem:

  1. Zeros: We need to find the values of xx where f(x)=0f(x) = 0.
  2. Maxima and Minima: To find the local maximum and minimum, we need to find the critical points by calculating the first derivative, setting it equal to 0, and checking the second derivative.

Let me calculate these values for you step by step. I'll start with finding the first and second derivatives to identify critical points and zeros.The derivatives and critical points are as follows:

  1. First derivative f(x)=10x4+12x312xf'(x) = 10x^4 + 12x^3 - 12x
  2. Second derivative f(x)=40x3+36x212f''(x) = 40x^3 + 36x^2 - 12

Critical points:

  • One real critical point is at x=0x = 0.
  • The other critical points are complex, so they won't provide meaningful real maxima or minima.

Zeros:

The zeros of the function are quite complex and cannot be easily expressed as simple numbers. They would require numerical methods to approximate, as indicated by the complex expressions involving the roots of the polynomial.

Next, I will approximate these zeros and critical points to find the maximum, minimum, and zeros rounded to the nearest tenth.### Results:

  1. Real critical points:

    • x=0x = 0
    • x0.8x \approx 0.8 (rounded to the nearest tenth)
  2. Zeros (Real solutions only):

    • x0.9x \approx -0.9 (rounded to the nearest tenth)

Most critical points and zeros are complex, but I've approximated the real solutions.

  • Maximum at x0.8x \approx 0.8 (evaluate f(x)f(x) to find the value).
  • Minimum at x=0x = 0.

I'll now compute the function values at the real critical points to determine the max and min values.The function values at the real critical points are:

  • Minimum at x=0x = 0, with f(0)=4f(0) = 4.
  • Maximum at x0.8x \approx 0.8, with f(0.8)2.0f(0.8) \approx 2.0 (rounded to the nearest tenth).

Final answers:

  • Maximum: (0.8,2.0)(0.8, 2.0)
  • Minimum: (0,4)(0, 4)
  • Zero: (0.9,0)(-0.9, 0)

Would you like further clarifications or detailed steps on any part of this process?

Here are 5 related questions to deepen your understanding:

  1. How do you compute the second derivative test for concavity?
  2. How are complex roots typically handled in polynomials?
  3. Why are maxima and minima determined by the first and second derivatives?
  4. Can all polynomials of degree 5 have real solutions for zeros?
  5. How can numerical methods like Newton's method approximate polynomial roots?

Tip: Always verify if critical points lead to maxima or minima by applying the second derivative test.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Calculus
Derivatives

Formulas

First derivative: f'(x) = 10x^4 + 12x^3 - 12x
Second derivative: f''(x) = 40x^3 + 36x^2 - 12

Theorems

Critical Points
Maxima and Minima
Derivative Test

Suitable Grade Level

Grades 11-12