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Secant of a Curve y =f(x)

The abscissa ...

Question

The abscissa of the points, where the tangent to curve $y = x^{3} - 3 x^{2} - 9 x + 5$ is parallel to x-axis, are

x = 0

and

0

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x = 1

and

- 1

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x = 1

and

- 3

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x = - 1

and

3

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Solution

The correct option is D $x = - 1$ and $3$
Given, $y = x^{3} - 3 x^{2} - 9 x + 5$
$\Rightarrow \frac{d y}{d x} = 3 x^{2} - 6 x - 9$
We know that, this equation gives the slope of the tangent to the curve. The tangent is parallel to x-axis,
$∴ \frac{d y}{d x} = 0$
$\Rightarrow 3 x^{2} - 6 x - 9 = 0$
$\Rightarrow x = - 1, 3$

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Similar questions

y = x^{3} - 3 x^{2} - 9 x + 7

y = (x^{3} - 1) (x - 2)

x -

y = x^{2} - x + 1

x_{1} = 0, x_{2} = - 1

x_{3} = \frac{5}{2}

y = x^{3} - 2 x^{2} - x

3 x - y + 1 = 0

y = x^{2} - x + 1,

x_{1} = 0, x_{2} = - 1

x_{3} = \frac{5}{2}

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