Find the point on the curve y = x³ – 11x + 5 at which the equation of tangent is y = x – 11.

Solution

The equation of the given curve is y = x³ - 11 x + 5.

The equation of the tangent to the given curve is given as y = x - 11 ( Which is of the form y = m x + c ).

$∴$ Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point ( x, y ) is

given by,

$\frac{dy}{dx} = 3 x^{2} - 11$

Then, we have:

3 x² - 11 = 1

$\Rightarrow 3 x^{2} = 12 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

When x = 2, y = ( 2 )³ - 11 ( 2 ) + 5

= 8 - 22 + 5

= - 9.

When x = - 2, y = ( - 2 )³ - 11 ( - 2 ) + 5

= - 8 + 22 + 5

= 19.

Hence, the required points are ( 2, - 9 ) and ( - 2, 19 ).

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