Calculating Horizontal Tangents for f(x)

Problem Statement

Find the x-coordinate of all points on the graph of the function
f(x) = x^3 - x^2 - x + 5
at which the tangent line is horizontal.

A tangent line to a curve is horizontal when the slope of the curve at that point is zero.
To find the slope of the curve, we need to determine the first derivative of the function, f'(x).

We compute the derivative of the function using the power rule, which states that the derivative of x^n is n \cdot x^{n-1}.
Therefore,
f'(x) = 3x^2 - 2x - 1.

To find the x-coordinates where the tangent line is horizontal, we set the derivative equal to zero:
3x^2 - 2x - 1 = 0.

We will apply the quadratic formula where for a general equation ax^2 + bx + c = 0, the roots are given by:
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
In our equation:
- a = 3
- b = -2
- c = -1
Plugging in these values into the quadratic formula:
x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}
- Calculating the discriminant:
- (-2)^2 = 4
- -4 \cdot 3 \cdot -1 = 12
- Therefore,
  4 + 12 = 16
- Now substituting back:
  x = \frac{2 \pm \sqrt{16}}{6}
- Which simplifies to:
  x = \frac{2 \pm 4}{6}
- This gives us two possible solutions:
1. x = \frac{6}{6} = 1
2. x = \frac{-2}{6} = -\frac{1}{3}.

The x-coordinates where the tangent line is horizontal are:
- (A) x = 1
- (C) x = -\frac{1}{3}

The answer choices provided in the problem statement:
- (A) x = 1
- (B) x = -1
- (C) x = \frac{1}{3}
- (D) x = -3, 1
- (E) x = 1, -1
The correct answers based on the calculations are:
- (A) x = 1
- The choice of x = -\frac{1}{3} does not match any of the options listed above, indicating that there may be an error or omission in the answer choices.

The solutions indicate that the points on the graph of the function where the tangent line is horizontal are at
- x = 1 and x = -\frac{1}{3}.
Further steps should include verifying if any additional context or adjustments in the answer formatting is needed to check against provided answer choices.