Math Problem Statement

y=x^4-16x^2

Solution

Given the function y=x416x2y = x^4 - 16x^2, let's analyze it thoroughly.

1. Factor the expression:

We can simplify the expression by factoring it: y=x416x2=x2(x216)y = x^4 - 16x^2 = x^2(x^2 - 16) Notice that x216x^2 - 16 is a difference of squares: x216=(x4)(x+4)x^2 - 16 = (x - 4)(x + 4) Thus, the fully factored form is: y=x2(x4)(x+4)y = x^2 (x - 4)(x + 4)


2. Find the roots:

To find the roots, set y=0y = 0: x2(x4)(x+4)=0x^2 (x - 4)(x + 4) = 0 This gives:

  • x2=0x=0x^2 = 0 \quad \Rightarrow \quad x = 0
  • x4=0x=4x - 4 = 0 \quad \Rightarrow \quad x = 4
  • x+4=0x=4x + 4 = 0 \quad \Rightarrow \quad x = -4

Thus, the roots are: x=4,x=0,x=4x = -4, \quad x = 0, \quad x = 4


3. Determine the behavior around the roots (sign analysis):

For large positive or negative xx, the x4x^4-term dominates, so the function grows rapidly in both directions. Now, let's analyze the sign of yy around the critical points x=4,0,4x = -4, 0, 4.

  • Interval (,4)(-\infty, -4): All terms are positive, so y>0y > 0.
  • Interval (4,0)(-4, 0): The factor (x+4)(x + 4) is positive, but (x4)(x - 4) is negative, so y<0y < 0.
  • Interval (0,4)(0, 4): The factor x2x^2 is positive, but (x4)(x - 4) is negative, so y<0y < 0.
  • Interval (4,)(4, \infty): All terms are positive, so y>0y > 0.

4. Critical points and turning points:

To find critical points, take the derivative of y=x416x2y = x^4 - 16x^2.

dydx=4x332x\frac{dy}{dx} = 4x^3 - 32x Factor the derivative: dydx=4x(x28)=4x(x22)(x+22)\frac{dy}{dx} = 4x(x^2 - 8) = 4x(x - 2\sqrt{2})(x + 2\sqrt{2}) Setting the derivative to zero: 4x(x22)(x+22)=04x(x - 2\sqrt{2})(x + 2\sqrt{2}) = 0 The critical points are: x=0,x=22,x=22x = 0, \quad x = 2\sqrt{2}, \quad x = -2\sqrt{2}


5. Second derivative for concavity:

Now, take the second derivative of yy: d2ydx2=12x232\frac{d^2y}{dx^2} = 12x^2 - 32 Set this equal to 0 to find inflection points: 12x232=012x2=32x2=8312x^2 - 32 = 0 \quad \Rightarrow \quad 12x^2 = 32 \quad \Rightarrow \quad x^2 = \frac{8}{3} Thus, the inflection points occur at: x=±263x = \pm \frac{2\sqrt{6}}{3}


6. Graph behavior summary:

  • Roots: x=4,0,4x = -4, 0, 4
  • Critical points: x=0,±22x = 0, \pm 2\sqrt{2}
  • Inflection points: x=±263x = \pm \frac{2\sqrt{6}}{3}
  • The function tends towards ++\infty as x±x \to \pm \infty.

Would you like a detailed plot or further explanation on any step?


Related questions:

  1. How does the second derivative test determine concavity?
  2. How do critical points relate to maxima or minima?
  3. What is the significance of inflection points in graphing?
  4. How does sign analysis help determine intervals of positivity and negativity?
  5. Can you describe the symmetry of this function?

Tip: When graphing, always look at end behavior to determine how the function behaves as xx \to \infty or xx \to -\infty.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Functions
Factoring
Critical Points
Sign Analysis
Concavity

Formulas

y = x^4 - 16x^2 = x^2(x - 4)(x + 4)
First derivative: dy/dx = 4x^3 - 32x
Second derivative: d^2y/dx^2 = 12x^2 - 32

Theorems

Difference of Squares
Critical Points
Second Derivative Test

Suitable Grade Level

Grades 11-12