1.

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

Answer»

Given equation of curve is y = x3 − 11x + 5.

The slope of tangent to the curve at point (x, y) is 3x2 − 11. ... (1)

Given that the equation of tangent to the curve at point (x, y) is y = x − 11.

Therefore, the slope of tangent to the curve at point (x, y) is 1. ... (2)

Therefore, 3x2 − 11 = 1

⇒ 3x2 = 12

⇒ x2 = 4

⇒ x = ±2.

If x = 2 then, y = 2 − 11 = −9

which also satisfies the equation of given curve.

And x = −2, then y = −2 − 11 = −13 which not satisfying the equation of curve y = x 3 − 11x + 5 .

Therefore, (−2, −13) is not the point where the tangent to the curve = x3 − 11x + 5 is = x − 11.

Hence, the tangent to the curve y = x3 − 11x + 5 at the point (2, -9) is y = x − 11.



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