{"id":6794,"date":"2021-10-21T17:11:43","date_gmt":"2021-10-21T11:41:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6794"},"modified":"2021-10-24T23:54:30","modified_gmt":"2021-10-24T18:24:30","slug":"find-the-inverse-of-the-function-fx-log_ax-sqrtx21-a-1-and-assuming-it-to-be-an-onto-function","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-inverse-of-the-function-fx-log_ax-sqrtx21-a-1-and-assuming-it-to-be-an-onto-function\/","title":{"rendered":"Find the inverse of the function f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\); a > 1 and assuming it to be an onto function."},"content":{"rendered":"

Solution :<\/h2>\n

Given f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\)<\/p>\n

f'(x) = \\(log_ae\\over {\\sqrt{1+x^2}}\\) > 0<\/p>\n

which is strictly increasing functions.<\/p>\n

Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible.<\/p>\n

Interchanging x & y<\/p>\n

\\(\\implies\\)\u00a0 \\(log_a(y + \\sqrt{(y^2+1)})\\) = x<\/p>\n

\\(\\implies\\)\u00a0 \\(y + \\sqrt{(y^2+1)}\\) = \\(a^x\\) ……..(1)<\/p>\n

and\u00a0 \\(\\sqrt{(y^2+1)}\\) – y = \\(a^{-x}\\) ………..(2)<\/p>\n

From (1) and (2), we get y = \\(1\\over 2\\)(\\(a^x – a^{-x}\\)) or \\(f{-1}\\)(x) = \\(1\\over 2\\)(\\(a^x – a^{-x}\\)).<\/p>\n

\n

Similar Questions<\/h3>\n

If y = 2[x] + 3 & y = 3[x \u2013 2] + 5, then find [x + y] where [.] denotes greatest integer function.<\/a><\/p>\n

Find the domain and range of function f(x) = \\(x-2\\over 3-x\\).<\/a><\/p>\n

Find the period of the function f(x) = \\(e^{x-[x]+|cos\\pi x|+|cos2\\pi x|+ \u2026.. + |cosn\\pi x|}\\)<\/a><\/p>\n

Find the domain of the function f(x) = \\(1\\over x + 2\\).<\/a><\/p>\n

Find the range of the function \\(log_{\\sqrt{2}}(2-log_2(16sin^2x+1))\\)<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\) f'(x) = \\(log_ae\\over {\\sqrt{1+x^2}}\\) > 0 which is strictly increasing functions. Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Interchanging x & y \\(\\implies\\)\u00a0 \\(log_a(y + \\sqrt{(y^2+1)})\\) = x \\(\\implies\\)\u00a0 \\(y + \\sqrt{(y^2+1)}\\) = \\(a^x\\) ……..(1) and\u00a0 \\(\\sqrt{(y^2+1)}\\) – …<\/p>\n

Find the inverse of the function f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\); a > 1 and assuming it to be an onto function.<\/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":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[50,43],"tags":[],"yoast_head":"\nFind the inverse of the function f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\); a > 1 and assuming it to be an onto function.<\/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\/find-the-inverse-of-the-function-fx-log_ax-sqrtx21-a-1-and-assuming-it-to-be-an-onto-function\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the inverse of the function f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\); a > 1 and assuming it to be an onto function.\" \/>\n<meta property=\"og:description\" content=\"Solution : Given f(x) = (log_a(x + sqrt{(x^2+1)})) f'(x) = (log_aeover {sqrt{1+x^2}}) > 0 which is strictly increasing functions. Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Interchanging x & y (implies)\u00a0 (log_a(y + sqrt{(y^2+1)})) = x (implies)\u00a0 (y + sqrt{(y^2+1)}) = (a^x) ……..(1) and\u00a0 (sqrt{(y^2+1)}) – … Find the inverse of the function f(x) = (log_a(x + sqrt{(x^2+1)})); a > 1 and assuming it to be an onto function. 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