Math Problem Statement

Find the point(s), if any, at which the graph of f has a horizontal tangent line. f(x) = x2 x − 7 (x, y) =

smaller x-value (x, y) =

larger x-value

Solution

To determine the points at which the graph of the function f(x)=x2x7f(x) = \frac{x^2}{x - 7} has a horizontal tangent line, we need to find where the derivative of the function equals zero. A horizontal tangent line occurs when the derivative of the function f(x)f'(x) is equal to zero.

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

The function is a quotient, so we will apply the quotient rule to differentiate it. The quotient rule states that:

f(x)=g(x)h(x)g(x)h(x)[h(x)]2f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}

Where:

  • g(x)=x2g(x) = x^2
  • h(x)=x7h(x) = x - 7

Now, differentiate both g(x)g(x) and h(x)h(x):

  • g(x)=2xg'(x) = 2x
  • h(x)=1h'(x) = 1

Thus, the derivative of f(x)f(x) is:

f(x)=(2x)(x7)(x2)(1)(x7)2f'(x) = \frac{(2x)(x - 7) - (x^2)(1)}{(x - 7)^2}

Simplify the numerator:

f(x)=2x(x7)x2(x7)2f'(x) = \frac{2x(x - 7) - x^2}{(x - 7)^2} f(x)=2x214xx2(x7)2f'(x) = \frac{2x^2 - 14x - x^2}{(x - 7)^2} f(x)=x214x(x7)2f'(x) = \frac{x^2 - 14x}{(x - 7)^2}

Step 2: Find where the derivative equals zero

To find the points where the graph has a horizontal tangent, we set the derivative equal to zero:

f(x)=0    x214x(x7)2=0f'(x) = 0 \implies \frac{x^2 - 14x}{(x - 7)^2} = 0

For the fraction to be zero, the numerator must be zero (since the denominator can never be zero unless x=7x = 7, which is not in the domain of the function).

So, set the numerator equal to zero:

x214x=0x^2 - 14x = 0 Factor the equation:

x(x14)=0x(x - 14) = 0

Thus, the solutions are:

x=0orx=14x = 0 \quad \text{or} \quad x = 14

Step 3: Find the corresponding yy-values

Now, we need to find the corresponding yy-values for x=0x = 0 and x=14x = 14 by substituting these xx-values into the original function f(x)=x2x7f(x) = \frac{x^2}{x - 7}.

  • When x=0x = 0:

f(0)=0207=07=0f(0) = \frac{0^2}{0 - 7} = \frac{0}{-7} = 0

So, the point is (0,0)(0, 0).

  • When x=14x = 14:

f(14)=142147=1967=28f(14) = \frac{14^2}{14 - 7} = \frac{196}{7} = 28

So, the point is (14,28)(14, 28).

Final Answer:

The points where the graph has horizontal tangent lines are:

  • (0,0)(0, 0) (smaller x-value)
  • (14,28)(14, 28) (larger x-value)

Would you like further details or explanations about this solution?

Related Questions:

  1. How would you apply the quotient rule in other contexts?
  2. What does the second derivative test tell us about the nature of the critical points?
  3. Can horizontal tangents occur in all types of functions?
  4. How does the behavior of f(x)f(x) change around the points of horizontal tangency?
  5. What is the relationship between horizontal tangent lines and local extrema?

Tip: For functions involving quotients, practicing the quotient rule will help solidify your understanding of how to differentiate more complex functions.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Derivative
Horizontal Tangent Line
Quotient Rule

Formulas

Quotient Rule: f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2

Theorems

Horizontal Tangent Line Theorem

Suitable Grade Level

Grades 11-12