Question

The equation of the tangent to the curve \(y = 3x^3-3x^2+x\) at \(x = 1\) is

• A. \(4x-y+3 = 0\)
• B. \(4x+y-3 = 0\)
• C. \(4x-y-3 = 0\)
• D. \(4x+y+3 = 0\)

Hint:

Find the point on the curve at x = 1 and the slope dy/dx there, then write y - y1 = m(x - x1).

Answer 1

The Correct Option is C

Step 1: Check each option for the point:
Put $(1,1)$ in each line. (A): $4-1+3=6\neq0$. (B): $4+1-3=2\neq0$. (C): $4-1-3=0$, yes. (D): $4+1+3=8\neq0$.

Step 2: Confirm the slope:
Only (C) passes through the point, so check its slope: $4x-y-3=0$ gives $y=4x-3$, slope 4.

Step 3: Compare with the curve:
The derivative at $x=1$ is $9-6+1=4$, which equals the slope of line (C).

Final Answer:
The tangent line is $4x-y-3=0$ (option C). \[ \boxed{4x-y-3=0} \]