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