Question

The abscissa of the points, where the tangent to the curve $y = x^3 - 3x^2 - 9x + 5$ is parallel to $X$ axis are

• A. $x = 1$ and $-1$
• B. $x = 1$ and $-3$
• C. $x = -1$ and $3$
• D. $x = 0$ and $1$

Hint:

"Parallel to the $X$-axis" is a geometric code phrase for setting the derivative equal to zero ($\frac{dy}{dx} = 0$).
"Parallel to the $Y$-axis" means the tangent is vertical, so the derivative is undefined (set the denominator of $\frac{dy}{dx}$ equal to zero).

Answer 1

The Correct Option is C

Step 1: Understand the question.
For the curve $y = x^3 - 3x^2 - 9x + 5$ we want the $x$ values where the tangent is parallel to the $X$-axis.
Step 2: Link tangent to slope.
The slope of the tangent is $\dfrac{dy}{dx}$. A line parallel to the $X$-axis is flat, so its slope is 0. We set the derivative to 0.
Step 3: Differentiate the curve.
\[ \frac{dy}{dx} = 3x^2 - 6x - 9 \]
Step 4: Set the slope to zero.
\[ 3x^2 - 6x - 9 = 0 \] Divide by 3 to make it simpler: \[ x^2 - 2x - 3 = 0 \]
Step 5: Factor the quadratic.
Split the middle term: $x^2 - 3x + x - 3 = 0$, so $x(x-3) + 1(x-3) = 0$, giving \[ (x-3)(x+1) = 0 \]
Step 6: Read the roots.
\[ x = 3 \quad \text{or} \quad x = -1 \] These are the required abscissae. \[ \boxed{x = -1 \text{ and } 3} \]