{"id":8144,"date":"2021-11-14T17:35:33","date_gmt":"2021-11-14T12:05:33","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8144"},"modified":"2021-11-15T16:03:44","modified_gmt":"2021-11-15T10:33:44","slug":"find-the-point-on-the-curve-y-cos-x-1-x-in-piover-2-3piover-2-at-which-tangent-is-parallel-to-the-x-axis","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-point-on-the-curve-y-cos-x-1-x-in-piover-2-3piover-2-at-which-tangent-is-parallel-to-the-x-axis\/","title":{"rendered":"Find the point on the curve y = cos x – 1, x \\(\\in\\) \\([{\\pi\\over 2}, {3\\pi\\over 2}]\\) at which tangent is parallel to the x-axis."},"content":{"rendered":"
Let f(x) = cos x – 1, Clearly f(x) is continous on \\([{\\pi\\over 2}, {3\\pi\\over 2}]\\) and differentiable on \\(({\\pi\\over 2}, {3\\pi\\over 2})\\).<\/p>\n
Also, f\\((\\pi\\over 2)\\) = \\(cos {\\pi\\over 2}\\) – 1 = -1 = f\\((3\\pi\\over 2)\\).<\/p>\n
Thus, all the conditions of rolle’s theorem<\/a> are satisfied. Consequently,there exist at least one point c \\(\\in\\) \\(({\\pi\\over 2}, {3\\pi\\over 2})\\) for which f'(c) = 0. But,<\/p>\n f'(c) = 0 \\(\\implies\\)\u00a0 -sin c = 0\u00a0 \\(\\implies\\)\u00a0 c = \\(\\pi\\)<\/p>\n \\(\\therefore\\)\u00a0 \u00a0f(c) = \\(cos \\pi\\) – 1 = -2<\/p>\n By the geometric interpretation of rolle’s theorem<\/a> (\\(\\pi\\), -2) is the point on y = cos x – 1 where tangent is parallel to x-axis.<\/p>\n Find the approximate value of f(3.02), where f(x) = \\(3x^2 + 5x + 3\\).<\/a><\/p>\n Verify Rolle\u2019s theorem for the function f(x) = \\(x^2\\) \u2013 5x + 6 on the interval [2, 3].<\/a><\/p>\n It is given that for the function f(x) = \\(x^3 \u2013 6x^2 + ax + b\\) on [1, 3], Rolles\u2019s theorem holds with c = \\(2 +{1\\over \\sqrt{3}}\\). Find the values of a and b, if f(1) = f(3) = 0.<\/a><\/p>\n
\nSimilar Questions<\/h3>\n