91Ó°ÊÓ

First, let's find the domains for each of the function's branches: 1. For \(x\leq1\), \(f(x) = 1-2x-x^2\). 2. For \(x>1\), \(f(x) = 9-6x+x^2\). To find the intervals where the function is monotonic, we will find the critical points of each branch by taking their first derivatives and setting them to zero. A function is monotonic if its first derivative has a constant sign throughout the interval (either always positive for an increasing function, or negative for a decreasing function). 1. For \(x\leq1\), \(f'(x) = -2-2x\). Setting \(f'(x)=0\) gives \(x=-1\). Since the first derivative is always negative for \(x\leq1\), \(f(x)\) is decreasing on the interval \((-\infty, 1]\). 2. For \(x>1\), \(f'(x) = -6+2x\). Setting \(f'(x)=0\) gives \(x=3\). Since the first derivative is always positive for \(x>1\), \(f(x)\) is increasing on the interval \((1, +\infty)\). Therefore, the function is monotonic on the intervals \((-\infty, 1]\) and \((1,+\infty)\).

Now we will find the inverse functions for each interval: 1. For \(x \leq 1\), the function is \(f(x) = 1-2x-x^2\). To find its inverse, let \(x = 1-2y-y^2\). Then solve for \(y\): $$ x = 1-2y-y^2 \\ y^2 + 2y + (1-x) = 0. $$ Solving this quadratic equation for \(y\), we get the inverse function \(f^{-1}(x) = -1+\sqrt{2-x}\). 2. For \(x > 1\), the function is \(f(x) = 9-6x+x^2\). To find its inverse, let \(x = 9-6y+y^2\). Then solve for \(y\): $$ x = 9-6y+y^2 \\ y^2 -6y +(9-x)=0. $$ Solving this quadratic equation for \(y\), we get the inverse function \(f^{-1}(x) = 3-\sqrt{10-x}\).

To sketch the functions, follow these steps: 1. For the original function \(f(x)\), mark the two points where the function changes its expression: \((1, 1-2(1)-(1)^2)=(1,-2)\) and \((1,9-6(1)+(1)^2)=(1,4)\). 2. Sketch the parabola \(f(x) = 1-2x-x^2\) for \(x\leq1\), which is a downward-opening parabola. 3. Sketch the parabola \(f(x) = 9-6x+x^2\) for \(x>1\), which is an upward-opening parabola. 4. Now, sketch the inverse function \(f^{-1}(x) = -1+\sqrt{2-x}\) for \(x\leq1\). It will be a reflection of \(f(x)\) for \(x\leq1\) with respect to the line \(y=x\). 5. Finally, sketch the inverse function \(f^{-1}(x) = 3-\sqrt{10-x}\) for \(x>1\). It will be a reflection of \(f(x)\) for \(x>1\) with respect to the line \(y=x\). The sketch should show the original function \(f(x)\) and its inverses \(f^{-1}(x)\) for their respective intervals.