Question

The equation of the tangent to the curve \(y = x^3\) at (1, 1) is

• A. \(3x - y + 2 = 0\)
• B. \(x - 10y - 50 = 0\)
• C. \(3x - y - 2 = 0\)
• D. \(x - 10y + 50 = 0\)

Hint:

To quickly check your answer, ensure two things: 1. The point (1,1) must satisfy the final equation. For option (C), \(3(1) - 1 - 2 = 3-3 = 0\). It works. 2. The slope of the line from the equation must match your calculated slope. For \(3x - y - 2 = 0\), the slope is \(-A/B = -3/(-1) = 3\), which matches.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find x where f(x)=2x³+3x²-36x+10 has local minimum.

Step 2: Key Formula (Alternate):
First derivative test: f'(x)=0 gives critical points. Second derivative: f''(x)>0 → minimum.

Step 3: Detailed Explanation:
f'(x)=6x²+6x-36=0 → x²+x-6=0 → x=-3,2. f''(x)=12x+6. f''(-3)=-30<0 (max). f''(2)=30>0 (min).

Step 4: Final Answer:
Minimum at x=2.