91Ó°ÊÓ

We are given the function \(h(t)=\frac{1+t}{2-t}\). Let's replace \(h(t)\) with \(x\): \[x = \frac{1+t}{2-t}\]

Now, we'll swap the roles of \(t\) and \(x\), so we get: \[t = \frac{1+x}{2-x}\]

Next, we need to solve for \(x\): \[t = \frac{1+x}{2-x}\] \[(2-x)t = 1+x\] \[2t-xt = x+1\] \[2t-1= x(x+t)\] Now, we factor out \(x\) to get: \[x = \frac{2t-1}{t+1}\]

To check that this inverse function is correct, we'll evaluate \(h(h^{-1}(t))\) and \(h^{-1}(h(t))\) to make sure we obtain \(t\). 1. Evaluate \(h(h^{-1}(t))\): \[h\left(\frac{2t-1}{t+1}\right) = \frac{1+\frac{2t-1}{t+1}}{2-\frac{2t-1}{t+1}}\] Clear out the fraction by multiplying both the numerator and denominator with \((t+1)\): \[\frac{(t+1)(1+\frac{2t-1}{t+1})}{(t+1)(2-\frac{2t-1}{t+1})}\] \[\frac{(t+1)(1+2t-1)}{(t+1)(2-2t+1)}\] \[\frac{t(1+2t)}{t(1-2t)} = t\] 2. Evaluate \(h^{-1}(h(t))\): \[h^{-1}\left(\frac{1+t}{2-t}\right) = \frac{2\left(\frac{1+t}{2-t}\right)-1}{\left(\frac{1+t}{2-t}\right)+1}\] Clear out the fraction by multiplying both the numerator and denominator with \((2-t)\): \[\frac{(2-t)(2\left(\frac{1+t}{2-t}\right)-1)}{(2-t)\left(\left(\frac{1+t}{2-t}\right)+1\right)}\] \[\frac{(2-t)(2+2t-1)}{(2-t)(1+t+2-1)}\] \[\frac{(2-t)(1+2t)}{(2-t)(1+t)} = t\] Since both \(h(h^{-1}(t))\) and \(h^{-1}(h(t))\) equal \(t\), our inverse function \(h^{-1}(t) = \frac{2t-1}{t+1}\) is indeed correct.