Math Problem Statement

For which values of x is the tangent line to the graph of f(x) = (3x^2 - 2)(2x + 1) horizontal?

Solution

To find the values of xx where the tangent line to the graph of f(x)=(3x22)(2x+1)f(x) = (3x^2 - 2)(2x + 1) is horizontal, we need to follow these steps:

Steps:

  1. Find the derivative of f(x)f(x):

    • The slope of the tangent line is given by the derivative f(x)f'(x).
    • If the tangent line is horizontal, the slope of the tangent line is 0, i.e., f(x)=0f'(x) = 0.
  2. Use the product rule to differentiate f(x)=(3x22)(2x+1)f(x) = (3x^2 - 2)(2x + 1):

    • The product rule is given by: ddx[u(x)v(x)]=u(x)v(x)+u(x)v(x)\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
    • Let u(x)=3x22u(x) = 3x^2 - 2 and v(x)=2x+1v(x) = 2x + 1.
  3. Differentiate each term:

    • u(x)=ddx(3x22)=6xu'(x) = \frac{d}{dx}(3x^2 - 2) = 6x
    • v(x)=ddx(2x+1)=2v'(x) = \frac{d}{dx}(2x + 1) = 2
  4. Apply the product rule: f(x)=(6x)(2x+1)+(3x22)(2)f'(x) = (6x)(2x + 1) + (3x^2 - 2)(2) Simplify: f(x)=6x(2x+1)+2(3x22)f'(x) = 6x(2x + 1) + 2(3x^2 - 2) f(x)=12x2+6x+6x24f'(x) = 12x^2 + 6x + 6x^2 - 4 f(x)=18x2+6x4f'(x) = 18x^2 + 6x - 4

  5. Set f(x)=0f'(x) = 0 and solve for xx: 18x2+6x4=018x^2 + 6x - 4 = 0 Divide by 2: 9x2+3x2=09x^2 + 3x - 2 = 0 Solve this quadratic equation using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=9a = 9, b=3b = 3, and c=2c = -2. x=3±324(9)(2)2(9)x = \frac{-3 \pm \sqrt{3^2 - 4(9)(-2)}}{2(9)} x=3±9+7218x = \frac{-3 \pm \sqrt{9 + 72}}{18} x=3±8118x = \frac{-3 \pm \sqrt{81}}{18} x=3±918x = \frac{-3 \pm 9}{18}

    So, the two solutions are: x=3+918=618=13x = \frac{-3 + 9}{18} = \frac{6}{18} = \frac{1}{3} x=3918=1218=23x = \frac{-3 - 9}{18} = \frac{-12}{18} = \frac{-2}{3}

Conclusion:

The tangent line is horizontal when x=13x = \frac{1}{3} and x=23x = \frac{-2}{3}.

Would you like more details or explanations about any step?

Here are 5 related questions for further exploration:

  1. How do you apply the product rule for differentiation?
  2. What are the conditions for a tangent line to be horizontal?
  3. How can you solve quadratic equations using the quadratic formula?
  4. What is the significance of the discriminant in the quadratic formula?
  5. How can the behavior of a function’s derivative inform us about its graph?

Tip: When solving quadratic equations, always check if the discriminant (the part under the square root) is positive, zero, or negative to understand the number of real solutions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Quadratic Equations

Formulas

Product Rule for Derivatives: d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
Quadratic Equation: ax^2 + bx + c = 0
Quadratic Formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a

Theorems

Product Rule for Differentiation
Quadratic Formula

Suitable Grade Level

Grades 10-12