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Find points where tangent line for curve f(x) = x^3-3x+2 is parallel with line 3x?
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You need to use the relation between the slopes of two parallel lines such that:
`m_1 = m_2`
Identifying the slope of the tangent line to the given curve at a point `(x_0,y_0)` yields:
`m_1 = f'(x_0)`
Identifying the slope of the parallel line to the tangent line, yields:
`m_2 = 3`
Equating `m_1 = m_2` yields:
`f'(x_0) = 3`
You need to evaluate derivative of the function at `x_0` , such that:
`f'(x_0) = (x_0^3-3x_0+2)' => f'(x_0) = 3x_0^2 - 3`
`3x_0^2 - 3 = 3 => 3x_0^2 = 6 => x_0^2 = 2 => x_0 = +-sqrt2`
Replacing `+-sqrt2` for `x_0` in equation `f(x_0) = (x_0^3-3x_0+2) ` yields:
`y_0 = f(sqrt 2) = 2sqrt2 - 3sqrt2 + 2 => y_0 = 2 - sqrt2`
`y_0 = f(-sqrt2) => f(-sqrt2) = -2sqrt2 + 3sqrt2 + 2 => y_0 = 2 + sqrt2`
Hence, evaluating the points where the tangent line to the curve` f(x) = x^3 - 3x + 2` is parallel to the line `f(x) = 3x` yields `x_0 = sqrt2, y_0 = 2 - sqrt2` and `x_0 = -sqrt 2, y_0 = 2 + sqrt2.`
Posted by sciencesolve on June 3, 2013 at 3:34 PM (Answer #1)
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