Question
Find the points on the curve y = `4x^3 - 3x + 5` at which the equation of the tangent is parallel to the x-axis.
Solution
`y = 4x^3 - 3x + 5 `...(i)
`dy/dx = 12x^2 - 3`
Given that lines is parallel to x-a xis
`:. dy/dx = 0`
`12x^2 - 3 = 0`
`12x^2 = 3`
`x^2 = 1/4`
`x= +- 1/2`
Put `x = +- 1/2` in equation (i)
`x = 1/2 "then " y = 4 (1/2)^3 - 3(1/2) + 5 = 4`
`:. "Point"(1/2, 4)`
when `x = (-1)/2` then `y = 4 ((-1)/(2)) - 3((-1)/2) + 5 = 6`
:. Point (x,y) = `(1/2, 4)` and `((-1)/2, 6)`
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Solution Find the Points on the Curve Y = `4x^3 - 3x + 5` at Which the Equation of the Tangent is Parallel to the X-axis. Concept: Tangents and Normals.
