Where does the curve y - x 4-2 x 2-2 have any horizontal tangents-A- 0-2-1-1 and -1-1B- 0-0-1-1 and -1-3C- 3-4-2-1 and -1-4D- 0-4-3-1 and -2-4

Given: y=x^4 2x^2+2 To get the tangents, Calculate dydx (slope of a tangent) ⇒dydx=d(x^4 2x^2+2)dx ⇒dydx =4x^3 4x+0 Now, dydx=0 ⇒ nbsp; 4x^3 4x=0 ⇒ ...

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Standard XII

Physics

Basic Differentiation Rule

Where does th...

Question

Where does the curve $y = x^{4} - 2 x^{2} + 2$ have any horizontal tangents?

(0, 2), (1, 1)

and

(- 1, 1)

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(0, 0), (- 1, 1)

and

(- 1, 3)

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(3, 4), (2, 1)

and

(- 1, 4)

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(0, 4), (3, 1)

and

(- 2, 4)

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Solution

The correct option is A $(0, 2), (1, 1)$ and $(- 1, 1)$
Calculate $\frac{d y}{d x}$ (slope of a tangent)
$\frac{d y}{d x} = \frac{d (x^{4} - 2 x^{2} + 2)}{d x}$
$\frac{d y}{d x} = 4 x^{3} - 4 x$
Solve the equation : $\frac{d y}{d x} = 0$ (since slope of horizontal tangent is zero)
$4 x^{3} - 4 x = 0$
$4 x (x^{2} - 1) = 0$
$x = 0, 1, - 1$
The curve $y = x^{4} - 2 x^{2} + 2$ have horizontal tangents at $x = 0, 1$ and $- 1$ . The corresponding points on the curve are $(0, 2), (1, 1)$ and $(- 1, 1)$ .

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Similar questions

Q. Find the coordinates of the points on the curve

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.
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Where does the curve y=x4−2x2+2 have any horizontal tangents?

Where does the curve $y = x^{4} - 2 x^{2} + 2$ have any horizontal tangents?