Let S be the set of all values of x for which the tangent to the curve y = f(x) = x³ - x² - 2x at (x

Q: Wait, why do we equate the derivative to the slope of the secant line?

That is the core of the problem! The derivative $f'(x)$ represents the slope of the tangent line at any point $x$. If the tangent is parallel to the secant line, they must have the same slope. By setting $f'(x)$ equal to the slope of the secant, we are mathematically enforcing the condition of parallelism.

Q: I got $x=1$ as one of the solutions. Does that mean the tangent at $x=1$ is the secant line itself?

Exactly! You have hit on a beautiful geometric insight. In this specific case, the secant line passes through the point $(1, f(1))$, and the tangent at that same point happens to have the same slope as the secant. This is perfectly valid; a line can be parallel to itself, and in this context, it is a degenerate case of the parallel condition.

Q: Is there a theorem that guarantees these points exist?

Yes! You are thinking of the Mean Value Theorem (MVT). MVT states that for a continuous and differentiable function on an interval $[a, b]$, there exists at least one point $c$ in $(a, b)$ where the tangent slope equals the secant slope. Our problem is essentially asking you to find those specific points $c$.

Question

Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f(x) = x^3 - x^2 - 2x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f(1))$ and $(-1, f(-1))$, then $S$ is equal to :

Options:

(A) $\{1/3, -1\}$

(B) $\{1/3, -1\}$

(C) $\{-1/3, 1\}$

(D) $\{1/3, 1\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The objective is to find the points on the cubic function $f(x) = x^3 - x^2 - 2x$ where the tangent line is parallel to the secant line connecting the points at $x=1$ and $x=-1$. This requires equating the instantaneous rate of change to the average rate of change over the given interval.

### Phase 1: The Secant Line

First, we determine the coordinates of the points $A$ and $B$ on the curve. Substituting $x=1$ into the function:
$$f(1) = (1)^3 - (1)^2 - 2(1) = 1 - 1 - 2 = -2$$
Thus, point $A$ is $(1, -2)$.

Next, we substitute $x=-1$ into the function:
$$f(-1) = (-1)^3 - (-1)^2 - 2(-1) = -1 - 1 + 2 = 0$$
Thus, point $B$ is $(-1, 0)$.

The slope $m$ of the secant line passing through $A(1, -2)$ and $B(-1, 0)$ is calculated as:
$$m = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{-2 - 0}{1 + 1} = \frac{-2}{2} = -1$$
This value, $m = -1$, represents the target slope for our tangent lines.

### Phase 2: The Derivative as the Bridge

To find the slope of the tangent at any point $x$, we calculate the derivative $f'(x)$. Applying the power rule to $f(x) = x^3 - x^2 - 2x$:
$$f'(x) = 3x^2 - 2x - 2$$

To find where the tangent is parallel to the secant, we set the derivative equal to the secant slope $m = -1$:
$$3x^2 - 2x - 2 = -1$$

### Phase 3: The Intersection

We now solve the resulting quadratic equation for $x$. Rearranging the terms gives:
$$3x^2 - 2x - 1 = 0$$

We factor the quadratic by splitting the middle term:
$$3x^2 - 3x + x - 1 = 0$$
$$3x(x - 1) + 1(x - 1) = 0$$
$$(3x + 1)(x - 1) = 0$$

Setting each factor to zero yields the solutions:
$$3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$$
$$x - 1 = 0 \Rightarrow x = 1$$

### Conclusion

We have identified the set of points $S = \{-\frac{1}{3}, 1\}$.

It is noteworthy that at $x=1$, the tangent line coincides with the secant line, while at $x=-\frac{1}{3}$, we find a distinct tangent line parallel to the secant. This result serves as a practical demonstration of the **Mean Value Theorem**, bridging the global average rate of change with the local instantaneous rate of change.

Animated Solution for Mathematics - Differentiation: Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f (x) = x^{3} - x^{2} - 2 x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f (1))$ and $(- 1, f (- 1))$ , then $S$ is equal to :

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Visualized Solution

Visualizing the Curve $f (x)$

Identifying the Secant Points

Calculating $f (1)$ and $f (- 1)$

Finding the Slope of the Secant

Slope Calculation

The Derivative as Slope

Calculating $f^{'} (x)$

Setting up the Equation

Simplifying the Quadratic

Factorizing the Expression

Solving for $x$

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

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Visualized Solution

Visualizing the Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)