1. **Problem:** Sketch the function $$f(x) = 2 \cos^{-1}(-x + 1) + 1$$ and find its domain and range. 2. **Recall the inverse cosine function properties:** - The function $$\cos^{-1}(z)$$ is defined for $$z \in [-1,1]$$. - Its range is $$[0, \pi]$$. 3. **Find the domain of $$f(x)$$:** Since $$f(x)$$ involves $$\cos^{-1}(-x + 1)$$, the argument must satisfy: $$-1 \leq -x + 1 \leq 1$$ 4. **Solve inequalities for $$x$$:** - From $$-1 \leq -x + 1$$: $$-1 \leq -x + 1 \implies -2 \leq -x \implies x \leq 2$$ - From $$-x + 1 \leq 1$$: $$-x + 1 \leq 1 \implies -x \leq 0 \implies x \geq 0$$ 5. **Domain:** $$0 \leq x \leq 2$$ 6. **Find the range of $$f(x)$$:** - Since $$\cos^{-1}(z) \in [0, \pi]$$, then: $$2 \cos^{-1}(-x + 1) \in [0, 2\pi]$$ - Adding 1 shifts the range: $$f(x) = 2 \cos^{-1}(-x + 1) + 1 \in [1, 2\pi + 1]$$ 7. **Summary:** - Domain: $$[0, 2]$$ - Range: $$[1, 2\pi + 1]$$ 8. **Graph behavior:** - At $$x=0$$, $$f(0) = 2 \cos^{-1}(1) + 1 = 2 \cdot 0 + 1 = 1$$ - At $$x=2$$, $$f(2) = 2 \cos^{-1}(-1) + 1 = 2 \pi + 1$$ This confirms the range.