Math Problem Statement

f(x)=1/10x^4-x^2-2

Solution

Let's analyze the given function f(x)=110x4x22f(x) = \frac{1}{10}x^4 - x^2 - 2. Here are some steps to better understand this function:

1. Key Features of the Function

  • Degree: This is a fourth-degree polynomial, so its graph will exhibit behavior typical of quartic functions, including the possibility of multiple turning points.
  • Leading Coefficient: The term 110x4\frac{1}{10}x^4 has a positive coefficient, so the graph will eventually rise as x±x \to \pm\infty.
  • Symmetry: Since all terms contain only even powers of xx, the function is symmetric about the yy-axis, meaning it is an even function.

2. Critical Points and Extrema

To find critical points, calculate the derivative and solve f(x)=0f'(x) = 0: f(x)=410x32x=25x32xf'(x) = \frac{4}{10}x^3 - 2x = \frac{2}{5}x^3 - 2x Factorize: f(x)=25x(x25)f'(x) = \frac{2}{5}x(x^2 - 5) Setting f(x)=0f'(x) = 0, we find: x=0orx=±5x = 0 \quad \text{or} \quad x = \pm\sqrt{5} These are the critical points.

3. Second Derivative Test

To determine whether the critical points correspond to maxima, minima, or points of inflection, compute the second derivative: f(x)=65x22f''(x) = \frac{6}{5}x^2 - 2 Evaluate f(x)f''(x) at the critical points:

  • At x=0x = 0: f(0)=2f''(0) = -2 (local maximum at x=0x = 0).
  • At x=±5x = \pm\sqrt{5}: f(±5)=65(5)2=62=4>0f''(\pm\sqrt{5}) = \frac{6}{5}(5) - 2 = 6 - 2 = 4 > 0 (local minima at x=±5x = \pm\sqrt{5}).

4. Intercepts

  • yy-Intercept: Set x=0x = 0: f(0)=2f(0) = -2 So the yy-intercept is (0,2)(0, -2).
  • xx-Intercepts: Solve f(x)=0f(x) = 0: 110x4x22=0\frac{1}{10}x^4 - x^2 - 2 = 0 Multiply through by 10: x410x220=0x^4 - 10x^2 - 20 = 0 Let u=x2u = x^2: u210u20=0u^2 - 10u - 20 = 0 Solve using the quadratic formula: u=(10)±(10)24(1)(20)2(1)=10±100+802=10±1802=10±652=5±35u = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(-20)}}{2(1)} = \frac{10 \pm \sqrt{100 + 80}}{2} = \frac{10 \pm \sqrt{180}}{2} = \frac{10 \pm 6\sqrt{5}}{2} = 5 \pm 3\sqrt{5} Since u=x2u = x^2, we have: x2=5+35andx2=535x^2 = 5 + 3\sqrt{5} \quad \text{and} \quad x^2 = 5 - 3\sqrt{5} Taking square roots gives four xx-intercepts: x=±5+35,x=±535x = \pm\sqrt{5 + 3\sqrt{5}}, \quad x = \pm\sqrt{5 - 3\sqrt{5}}

5. End Behavior

As x±x \to \pm\infty, the 110x4\frac{1}{10}x^4 term dominates, so f(x)+f(x) \to +\infty.

Would you like me to graph the function or go deeper into any of these steps? 😊


5 Related Questions:

  1. How do quartic functions behave differently from cubic functions?
  2. What are the implications of a function being even or odd?
  3. Can you explain the quadratic substitution method used here in detail?
  4. How do we confirm the symmetry of a function mathematically?
  5. How does the second derivative test help classify critical points?

Tip:

Always check the behavior of higher-order terms when analyzing polynomials, as they dictate the end behavior of the graph.

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Math Problem Analysis

Mathematical Concepts

Polynomial Functions
Critical Points
Symmetry
End Behavior

Formulas

Derivative: f'(x) = d/dx [f(x)]
Critical points: Solve f'(x) = 0
Second derivative: f''(x) = d^2/dx^2 [f(x)]
Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a

Theorems

Second Derivative Test
Even Function Symmetry

Suitable Grade Level

Grades 11-12