Question:medium
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:
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:
Show 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.
Updated On: May 2, 2026
- $(1, -4)$
- $(2, -9)$
- $(2, -4)$
- $(2, -1)$
- $(2, -5)$
Show Solution
The Correct Option is B
Solution and Explanation
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