Expert answers to "Find the points on the curve y=2x3+3x2−12x+1 where..."

ossidianaZ

Answered question

2021-09-13

Find the points on the curve

y = 2 x^{3} + 3 x^{2} - 12 x + 1

where the tangent is horizontal.

nitruraviX

Skilled2021-09-14Added 101 answers

We are looking for a solution to this equation:

y^{'} (x) = 0

because the tangent is horizontal when the derivative is zero.
So,

y^{'} (x) = 0

{(2 x^{3} + 3 x^{2} - 12 x + 1)}^{'} = 0

6 x^{2} + 6 x - 12 = 0

x^{2} + x - 2 = 0

Now we need to find a solution to this equation:

x^{2} + x - 2 = 0

This is a quadratic equation so the solution is given with this formula:

x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

We have

x_{1, 2} = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (- 2)}}{2 \cdot 1}

x_{1, 2} = \frac{- 1 \pm \sqrt{9}}{2}

x_{1} = \frac{- 1 + 3}{2}

x_{1} = 1

x_{2} = \frac{- 1 - 3}{2}

x_{2} = 2

So the given curve has horizontal tangent when x=1,-2, so the corresponding points are (1,-6) and (-2,21)
Result:
(1,-6), (-2,21)

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