<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1771918596901-483681bxzus.png" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div><div class="question-column" style="flex: 1; align-self: flex-start;"><p>• For&nbsp;\({\mathrm{f}^{-1}}\) to exist, \({\mathrm{f}}\) must be a one-one function.<br>• \({\mathrm{f}}\) is the reflection of \({\mathrm{f}^{-1}}\) in the line \({y = x}\).<br>• \({D_{\mathrm{f}^{-1}} = R_{\mathrm{f}}}\)<br>• \({R_{\mathrm{f}^{-1}} = D_{\mathrm{f}}}\)<br>• Point \({A(r, p)}\) on \({\mathrm{f}}\) will correspond to \({A'(p, r)}\) on \({\mathrm{f}^{-1}}\).<br>• To find \({\mathrm{f}(x) = \mathrm{f}^{-1}(x)}\) implies \({\mathrm{f}(x) = \mathrm{f}^{-1}(x) = x}\) as&nbsp;\({y = \mathrm{f}(x)}\), \({y = \mathrm{f}^{-1}(x)}\) and \({y = x}\) meet at the point(s) of intersections.</p></div></div></div>

Functions - Inverse Functions

Functions - Inverse Functions