1. **State the problem:** We are given two functions: $$f(x) = \sqrt{x} + 1$$ and $$g(x) = -x + 4$$. We want to understand these functions and possibly find their composition or intersection. 2. **Recall the formulas and rules:** - The square root function $$\sqrt{x}$$ is defined for $$x \geq 0$$. - The function $$g(x) = -x + 4$$ is a linear function with slope $$-1$$ and y-intercept $$4$$. 3. **Evaluate the functions at some points:** - For $$f(x)$$, since $$x \geq 0$$, for example, $$f(0) = \sqrt{0} + 1 = 1$$, $$f(4) = \sqrt{4} + 1 = 2 + 1 = 3$$. - For $$g(x)$$, $$g(0) = -0 + 4 = 4$$, $$g(4) = -4 + 4 = 0$$. 4. **Find the intersection points by solving $$f(x) = g(x)$$:** $$\sqrt{x} + 1 = -x + 4$$ Rearranged: $$\sqrt{x} = -x + 3$$ Since $$\sqrt{x} \geq 0$$, the right side must be non-negative: $$-x + 3 \geq 0 \implies x \leq 3$$ Square both sides: $$x = (-x + 3)^2 = (3 - x)^2 = 9 - 6x + x^2$$ Bring all terms to one side: $$0 = 9 - 6x + x^2 - x = 9 - 7x + x^2$$ Rewrite: $$x^2 - 7x + 9 = 0$$ 5. **Solve the quadratic equation:** Using the quadratic formula: $$x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 9}}{2} = \frac{7 \pm \sqrt{49 - 36}}{2} = \frac{7 \pm \sqrt{13}}{2}$$ 6. **Check domain restrictions:** Recall $$x \leq 3$$ from step 4. Calculate approximate values: $$\frac{7 - \sqrt{13}}{2} \approx \frac{7 - 3.606}{2} = \frac{3.394}{2} = 1.697$$ (valid) $$\frac{7 + \sqrt{13}}{2} \approx \frac{7 + 3.606}{2} = \frac{10.606}{2} = 5.303$$ (not valid since $$5.303 > 3$$) 7. **Find the corresponding $$y$$ value:** For $$x = \frac{7 - \sqrt{13}}{2}$$, $$y = f(x) = \sqrt{x} + 1$$ This is the intersection point. **Final answer:** The functions intersect at $$\left( \frac{7 - \sqrt{13}}{2}, \sqrt{\frac{7 - \sqrt{13}}{2}} + 1 \right)$$.