Question
Download Solution PDFFind the point at which the tangent to the curve y = x3 – 3x2 + 3x - 6 is parallel to the x-axis.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Slope of the curve = dy/dx
Calculation:
Let point on curve be (x, y)
Given: Equation of the curve y = x3 – 3x2 + 3x - 6 ---(1)
Differentiating with respect to x, we get
Since, the tangent is parallel to the axis,
⇒ 3x2 – 6x + 3 = 0
⇒ 3 (x2 – 2x + 1) = 0
⇒ (x – 1)2 = 0
∴ x = 1
Put the value of x in equation 1st,
⇒ y = 1 – 3 + 3 – 6 = -5
Therefore, the required points is (1, -5)
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