Find the points on the curve y=2x33x2−12x4 where the tangent line is horizontal

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rerCessbalmuh · Accepted Answer

Find the points on the curve where the tangent line is horizontal as follows.
The given curve is $y = 2 x^{3} + 3 x^{2} - 12 x + 4$
Compute the derivative of the function as follows.
$y ‘ = \frac{d}{d x} (2 x^{3} + 3 x^{2} - 12 x + 4)$
$= 6 x^{2} + 6 x - 12$
Equate the derivative equal to zero as follows.
$6 x^{2} + 6 x - 12 = 0$
$x^{2} + x - 2 = 0$
$(x + 2) (x - 1) = 0$
$x = - 1$ or $x = 1$
For the smaller value of x that is , the value of y is as follows.
$y = 2 {(- 2)}^{3} + 3 {(- 2)}^{2} - 12 (- 2) + 4$
$= - 16 + 12 + 24 + 4$
= 24
For the larger value of x that is x = 1, the value of y is as follows.
$y = 2 {(1)}^{3} + 3 {(1)}^{2} - 12 (1) + 4$
= -3
Thus, for the smaller value of x, (x, y) = (-2, 24) and for the larger values of x, (x, y) = (1, -3)

Find the points on the curve y = 2x^3 3x^2

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