1. **Stating the problem:** We are given two functions: - $f(x) = x^2 - 6x + 5$ - $g(x) = |f(-|x| + 1) - 2|$ We need to analyze the function $g(x)$ based on $f(x)$. 2. **Analyzing $f(x)$:** The function $f(x)$ is a quadratic. Let's rewrite it in vertex form by completing the square: $$f(x) = x^2 - 6x + 5 = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4$$ This tells us: - The vertex is at $(3, -4)$. - The parabola opens upwards. 3. **Understanding the input to $f$ in $g(x)$:** The input to $f$ in $g(x)$ is $- |x| + 1$. For any real $x$, $|x| \\geq 0$, so $- |x| + 1 \\leq 1$. This means the argument to $f$ ranges from $-\\infty$ to $1$, but since $|x|$ grows without bound, effectively the domain of input to $f$ in $g$ is $(-\\infty, 1]$. 4. **Evaluating $f(-|x|+1)$:** Substitute $u = -|x| + 1$, then $$f(u) = (u-3)^2 - 4$$ So, $$f(-|x| + 1) = ((- |x| + 1) - 3)^2 - 4 = (- |x| - 2)^2 - 4 = ( - |x| - 2)^2 - 4$$ 5. **Simplify the squared term:** Since $(-|x| - 2)^2 = (|x| + 2)^2 = |x|^2 + 4|x| + 4 = x^2 + 4|x| + 4$ Thus, $$f(-|x| + 1) = x^2 + 4|x| + 4 - 4 = x^2 + 4|x|$$ 6. **Form of $g(x)$:** Recall $g(x) = |f(-|x| +1) - 2|$, so $$g(x) = |x^2 + 4|x| - 2|$$ 7. **Interpreting $g(x)$:** The function takes the value of the quadratic-like expression with the absolute value inside and shifts it down by 2, then takes the absolute value of the result. 8. **Final expression:** $$\boxed{g(x) = |x^2 + 4|x| - 2|}$$ This expression fully describes $g$ in terms of $x$ without nested functions. **Summary:** We started with $$g(x) = |f(-|x| + 1) - 2|$$ where $$f(x) = x^2 - 6x + 5 = (x-3)^2 - 4$$ calculating and simplifying, we found $$g(x) = |x^2 + 4|x| - 2|$$