Question:
The equation of the tangent to the curve \(y = 3x^3-3x^2+x\) at \(x = 1\) is
The equation of the tangent to the curve \(y = 3x^3-3x^2+x\) at \(x = 1\) is
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Find the point on the curve at x = 1 and the slope dy/dx there, then write y - y1 = m(x - x1).
Updated On: Oct 1, 2026
- \(4x-y+3 = 0\)
- \(4x+y-3 = 0\)
- \(4x-y-3 = 0\)
- \(4x+y+3 = 0\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The tangent at a point has slope equal to \(\frac{dy}{dx}\) at that point. We need the point and the slope.
Step 2: Find the point:
At \(x=1\): \(y=3-3+1=1\). The point is \((1,1)\).
Step 3: Find the slope:
\(\frac{dy}{dx}=9x^2-6x+1\). At \(x=1\) this is \(9-6+1=4\).
Step 4: Write the equation:
Use the point-slope form \(y-y_1=m(x-x_1)\) with \(m=4\) and \((x_1,y_1)=(1,1)\).
\(y-1=4(x-1)\), so \(y=4x-3\), that is \(4x-y-3=0\).
Step 5: Why the other options are wrong:
Option (A) \(4x-y+3=0\) does not pass through \((1,1)\) because \(4-1+3=6\). Options (B) and (D) have slope \(-4\), so they are wrong. Only option (C) has slope 4 and passes through \((1,1)\).
Final Answer:
The tangent is \(4x-y-3=0\), option (C). \[ \boxed{4x-y-3=0} \]
The tangent at a point has slope equal to \(\frac{dy}{dx}\) at that point. We need the point and the slope.
Step 2: Find the point:
At \(x=1\): \(y=3-3+1=1\). The point is \((1,1)\).
Step 3: Find the slope:
\(\frac{dy}{dx}=9x^2-6x+1\). At \(x=1\) this is \(9-6+1=4\).
Step 4: Write the equation:
Use the point-slope form \(y-y_1=m(x-x_1)\) with \(m=4\) and \((x_1,y_1)=(1,1)\).
\(y-1=4(x-1)\), so \(y=4x-3\), that is \(4x-y-3=0\).
Step 5: Why the other options are wrong:
Option (A) \(4x-y+3=0\) does not pass through \((1,1)\) because \(4-1+3=6\). Options (B) and (D) have slope \(-4\), so they are wrong. Only option (C) has slope 4 and passes through \((1,1)\).
Final Answer:
The tangent is \(4x-y-3=0\), option (C). \[ \boxed{4x-y-3=0} \]
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