Math Problem Statement

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Find the point(s) on the graph of f(x) = x^3 - 12x + 1 for which the function has a horizontal tangent line.

Solution

We are tasked with finding the point(s) on the graph of the function f(x)=x312x+1f(x) = x^3 - 12x + 1 where the function has a horizontal tangent line.

Step 1: Understand the problem

  • A horizontal tangent line occurs when the slope of the tangent line is zero.
  • The slope of the tangent line is given by the derivative of the function f(x)f(x).

Step 2: Find the derivative of f(x)f(x)

We differentiate f(x)=x312x+1f(x) = x^3 - 12x + 1 with respect to xx:

f(x)=ddx(x3)ddx(12x)+ddx(1)f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(12x) + \frac{d}{dx}(1)

Using basic differentiation rules: f(x)=3x212f'(x) = 3x^2 - 12

Step 3: Set the derivative equal to zero

To find the points where the tangent line is horizontal, we need to set the derivative equal to zero and solve for xx:

3x212=03x^2 - 12 = 0

Step 4: Solve for xx

Simplify and solve the equation:

3x2=123x^2 = 12 x2=4x^2 = 4 x=±2x = \pm 2

So, the function has horizontal tangent lines at x=2x = 2 and x=2x = -2.

Step 5: Find the corresponding yy-values

To find the points, we substitute x=2x = 2 and x=2x = -2 into the original function f(x)=x312x+1f(x) = x^3 - 12x + 1:

  • For x=2x = 2: f(2)=2312(2)+1=824+1=15f(2) = 2^3 - 12(2) + 1 = 8 - 24 + 1 = -15

  • For x=2x = -2: f(2)=(2)312(2)+1=8+24+1=17f(-2) = (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17

Step 6: Conclusion

The points on the graph where the function has horizontal tangent lines are:

  • (2,15)(2, -15)
  • (2,17)(-2, 17)

Would you like any additional details or clarifications?


Here are 5 related questions to deepen your understanding:

  1. What is the geometric meaning of a horizontal tangent line in terms of the graph's shape?
  2. How do we determine whether the horizontal tangent line corresponds to a local maximum or minimum?
  3. How does the second derivative of a function help classify critical points (such as those with horizontal tangents)?
  4. Can a cubic function have more than two points with horizontal tangent lines?
  5. How would the function's behavior change if we altered the constants in f(x)=x312x+1f(x) = x^3 - 12x + 1?

Tip: When solving for critical points, always check the second derivative to classify the nature of the critical points (whether they are maxima, minima, or points of inflection).

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Math Problem Analysis

Mathematical Concepts

Differentiation
Critical Points
Algebra

Formulas

f'(x) = 3x^2 - 12
Set f'(x) = 0 to find critical points

Theorems

First Derivative Test

Suitable Grade Level

Grades 10-12