{"id":8017,"date":"2021-11-13T14:54:15","date_gmt":"2021-11-13T09:24:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8017"},"modified":"2021-11-13T16:24:21","modified_gmt":"2021-11-13T10:54:21","slug":"separate-0-piover-2-into-subintervals-in-which-fx-sin-3x-is-increasing-or-decreasing","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/separate-0-piover-2-into-subintervals-in-which-fx-sin-3x-is-increasing-or-decreasing\/","title":{"rendered":"Separate \\([0, {\\pi\\over 2}]\\) into subintervals in which f(x) = sin 3x is increasing or decreasing."},"content":{"rendered":"

Solution :<\/h2>\n

We have, f(x) = sin 3x<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0f'(x) = 3cos 3x<\/p>\n

Now,\u00a0 0 < x < \\(pi\\over 2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 0 < 3x < \\(3\\pi\\over 2\\)<\/p>\n

Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases.<\/p>\n

Case 1<\/strong> : When 0 < 3x < \\(\\pi\\over 2\\)\u00a0 i.e.\u00a0 0 < x < \\(\\pi\\over 6\\)<\/p>\n

In this case, we have<\/p>\n

0 < 3x < \\(\\pi\\over 2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 cos 3x > 0<\/p>\n

\\(\\implies\\)\u00a0 3 cos 3x > 0 \\(\\implies\\) f'(x) > 0<\/p>\n

Thus, f'(x) > 0 for 0 < 3x < \\(\\pi\\over 2\\)\u00a0 i.e.\u00a0 0 < x < \\(\\pi\\over 6\\)<\/p>\n

So, f(x) is increasing on \\((0, {\\pi\\over 6})\\).<\/p>\n

Case 2<\/strong> : When \\(\\pi\\over 2\\) < 3x < \\(3\\pi\\over 2\\)\u00a0 i.e.\u00a0 \\(\\pi\\over 6\\) < x < \\(\\pi\\over 2\\)<\/p>\n

In this case, we have<\/p>\n

\\(\\pi\\over 2\\) < 3x < \\(3\\pi\\over 2\\) \u00a0\\(\\implies\\)\u00a0 cos 3x < 0<\/p>\n

\\(\\implies\\)\u00a0 3 cos 3x < 0 \\(\\implies\\) f'(x) < 0<\/p>\n

Thus, f'(x) < 0 for \\(\\pi\\over 2\\) < 3x < \\(3\\pi\\over 2\\) \u00a0i.e.\u00a0 \\(\\pi\\over 6\\) < x < \\(\\pi\\over 2\\)<\/p>\n

So, f(x) is decreasing on \\(({\\pi\\over 6}, {\\pi\\over 2})\\).<\/p>\n

\n

Similar Questions<\/h3>\n

Prove that the function f(x) = \\(x^3 \u2013 3x^2 + 3x \u2013 100\\) is increasing on R<\/a><\/p>\n

Prove that \\(f(\\theta)\\) = \\({4sin \\theta\\over 2 + cos\\theta} \u2013 \\theta\\) is an increasing function of \\(\\theta\\) in \\([0, {\\pi\\over 2}]\\).<\/a><\/p>\n

Find the point of inflection for f(x) = \\(x^4\\over 12\\) \u2013 \\(5x^3\\over 6\\) + \\(3x^2\\) + 7.<\/a><\/p>\n

Find the point of inflection for the curve y = \\(x^3 \u2013 6x^2 + 12x + 5\\).<\/a><\/p>\n

Find the inflection point of f(x) = \\(3x^4 \u2013 4x^3\\).<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have, f(x) = sin 3x \\(\\therefore\\)\u00a0 \u00a0f'(x) = 3cos 3x Now,\u00a0 0 < x < \\(pi\\over 2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 0 < 3x < \\(3\\pi\\over 2\\) Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases. Case 1 : When 0 < …<\/p>\n

Separate \\([0, {\\pi\\over 2}]\\) into subintervals in which f(x) = sin 3x is increasing or decreasing.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[66,43],"tags":[],"yoast_head":"\nSeparate \\([0, {\\pi\\over 2}]\\) into subintervals in which f(x) = sin 3x is increasing or decreasing.<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/separate-0-piover-2-into-subintervals-in-which-fx-sin-3x-is-increasing-or-decreasing\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Separate \\([0, {\\pi\\over 2}]\\) into subintervals in which f(x) = sin 3x is increasing or decreasing.\" \/>\n<meta property=\"og:description\" content=\"Solution : We have, f(x) = sin 3x (therefore)\u00a0 \u00a0f'(x) = 3cos 3x Now,\u00a0 0 < x < (piover 2)\u00a0 \u00a0(implies)\u00a0 0 < 3x < (3piover 2) Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases. Case 1 : When 0 < … Separate ([0, {piover 2}]) into subintervals in which f(x) = sin 3x is increasing or decreasing. 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