Question

Let $f(x) = 2x^3 - 5x^2 - 4x + 3,\ \frac{1}{2} \leq x \leq 3$. The point at which the tangent to the curve is parallel to the x-axis is:

• A. $(1, -4)$
• B. $(2, -9)$
• C. $(2, -4)$
• D. $(2, -1)$
• E. $(2, -5)$

Hint:

"Parallel to the x-axis" is a common phrase in calculus that simply means $dy/dx = 0$. Always check your final x-values against the domain provided in the question to eliminate extraneous solutions.

Answer 1

The Correct Option is B

Concept: A tangent parallel to the x-axis has a slope of zero. Therefore, we must find the derivative of the function, set it to zero to find the critical x-values, and then determine the corresponding y-coordinate.

Step 1: Differentiate the function.
Given $f(x) = 2x^3 - 5x^2 - 4x + 3$: \[ f'(x) = 6x^2 - 10x - 4 \]

Step 2: Solve for $f'(x) = 0$.
Set the derivative to zero to find where the slope is zero: \[ 6x^2 - 10x - 4 = 0 \quad \Rightarrow \quad 2(3x^2 - 5x - 2) = 0 \] Factorizing the quadratic: \[ 3x^2 - 6x + x - 2 = 0 \quad \Rightarrow \quad 3x(x - 2) + 1(x - 2) = 0 \] \[ (3x + 1)(x - 2) = 0 \] The critical values are $x = -1/3$ and $x = 2$.

Step 3: Select the valid x-value and find the y-coordinate.
The interval is given as $\frac{1}{2} \leq x \leq 3$. Only $x = 2$ lies in this range. Substitute $x = 2$ back into $f(x)$: \[ f(2) = 2(2)^3 - 5(2)^2 - 4(2) + 3 \] \[ f(2) = 2(8) - 5(4) - 8 + 3 = 16 - 20 - 8 + 3 = -9 \] The point is $(2, -9)$.