f(x) = 1 - \cos^2 x + 2\cos x = -\cos^2 x + 2\cos x + 1.

["# Understanding f(x) = 1 – cos²x + 2cos x: A Comprehensive Guide", "SEO Meta Description:\nExplore the function f(x) = 1 – cos²x + 2cos x in depth — its algebraic form, simplification, graph, roots, derivatives, and applications. Perfect for students and math enthusiasts.", "---", "## Introduction to f(x) = 1 – cos²x + 2cos x", "In trigonometric algebra, simplifying compound expressions is essential for deeper understanding and easier problem-solving. One such function is:\nf(x) = 1 – cos²x + 2cos x\nThis expression combines polynomial terms with a cosine function, making it a key example for studying trigonometric identities, function behavior, and optimization.", "---", "## Simplifying f(x): Rewriting in Terms of cos x", "Let’s begin by rewriting f(x) using a substitution to simplify analysis. Let:\nu = cos x\nThen the function becomes:\nf(x) = 1 – u² + 2u", "This is now a quadratic expression in u:\nf(u) = –u² + 2u + 1", "The negative coefficient on (u^2) indicates this is a downward-opening parabola, which significantly affects its graph and extremum values.", "---", "### Completing the Square", "To better understand the function’s maximum and vertex, complete the square:\n[\nf(u) = -u^2 + 2u + 1 = -\left(u^2 - 2u\right) + 1\n]\nComplete the square inside the parentheses:\n[\nu^2 - 2u = (u - 1)^2 - 1\n]\nSubstitute back:\n[\nf(u) = -\left[(u - 1)^2 - 1\right] + 1 = -(u - 1)^2 + 1 + 1 = -(u - 1)^2 + 2\n]", "Thus,\nf(u) = –(u – 1)² + 2", "This reveals:\n- The maximum value of f(u) is 2, attained when (u = 1)\n- The vertex of the parabola is at (u = 1), so maximum output occurs when (\cos x = 1)\n- The expression decreases symmetrically as (u) moves away from 1", "---", "## Finding the Roots of f(x)", "To find when f(x) = 0:\n[\n1 – \cos^2 x + 2\cos x = 0\n]\nSubstitute (u = \cos x):\n[\n1 – u^2 + 2u = 0 \quad \Rightarrow \quad -u^2 + 2u + 1 = 0\n]\nMultiply through by –1:\n[\nu^2 – 2u – 1 = 0\n]\nSolve using the quadratic formula:\n[\nu = \frac{2 \pm \sqrt{(-2)^2 + 4(1)(1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\n]", "Now evaluate:\n- (1 + \sqrt{2} \approx 2.414 > 1) → Not valid (cosine values lie in [–1, 1])\n- (1 - \sqrt{2} \approx -0.414) → Valid", "Thus, the only real solution is:\n(\cos x = 1 - \sqrt{2})", "Then,\nx = \cos^{-1}(1 - \sqrt{2}) + 2\pi n or x = –\cos^{-1}(1 - \sqrt{2}) + 2\pi n, for any integer n", "This means the function crosses zero at specific x-values determined by this root—important for phase analysis and periodicity studies.", "---", "## The Maximum Value of f(x)", "From the vertex form:\n[\nf(u) = -(u - 1)^2 + 2\n]\nMaximum value is 2, achieved when (u = \cos x = 1), i.e., when (x = 2\pi n), (n \in \mathbb{Z})", "This maximum gives insight into the function’s range:\n[\n\boxed{-\infty < f(x) \leq 2}\n]\nEven though the quadratic opens downward, cosine constrains (u \in [-1, 1]), so the actual maximum within the domain is 2.", "---", "## Behavior of f(x) Over Its Domain", "- Domain: All real x where (\cos x) is defined → (x \in \mathbb{R})\n- Range: Since (\cos x \in [-1, 1]), evaluate f(u) at endpoints and vertex:\n - At (u = 1): (f = 1 – 1 + 2(1) = 2) (maximum)\n - At (u = -1): (f = 1 – 1 + 2(-1) = 1 – 1 – 2 = –2)\n - Vertex at (u = 1), already evaluated", "Thus, f(x) ranges from –2 up to 2, but only reaches 2 at specific points.", "---", "## Graphing f(x): Key Features", "- Shape: Downward-opening parabola in terms of (u = \cos x), but periodic due to trigonometric input\n- Periodicity: Governed by (\cos x), so f(x) has period (2\pi)\n- Extremes: Maximum = 2 at (\cos x = 1), minimum = –2 at (\cos x = -1)\n- Symmetry: Reflects cosine symmetry but reflects and squashes the parabola", "The graph oscillates between –2 and 2, peaking at (x = 2\pi n) and dipping at (x = \pi + 2\pi n), since (\cos(\pi) = -1).", "---", "## Derivative and Critical Points", "To analyze slopes and maxima, compute the derivative of f(x):\n[\nf(x) = 1 - \cos^2 x + 2\cos x \\nf'(x) = \frac{d}{dx}[1 - \cos^2 x + 2\cos x] = 2\cos x (-\sin x) + 2(-\sin x) = -2\cos x \sin x - 2\sin x\n]\nFactor:\n[\nf'(x) = -2\sin x (\cos x + 1)\n]", "Set derivative to zero:\n[\n-2\sin x (\cos x + 1) = 0\n]", "Solutions:\n1. (\sin x = 0 \Rightarrow x = n\pi), for integer n — critical points at even and odd multiples of π\n2. (\cos x + 1 = 0 \Rightarrow \cos x = -1 \Rightarrow x = \pi + 2\pi n)", "Evaluate f(x) at these points within one period (x \in [0, 2\pi]):\n- At (x = 0): (f(0) = 1 – 1 + 0 = 0)\n- At (x = \pi): (f(\pi) = 1 – 1 + 2(-1) = -2)\n- At (x = 2\pi): same as 0 → 0", "This confirms the earlier findings — minimum at (\cos x = -1), max at (\cos x = 1).", "---", "## Applications and Extensions", "Understanding f(x) supports several mathematical and applied contexts:", "- Signal Processing: Representing modulated cosine signals\n- Optimization: Maximizing expressions involving periodic inputs\n- Calculus Education: Teaching derivatives of composite functions and trigonometric simplification\n- Physics: Modeling oscillatory systems with phase shifts", "---", "## Summary", "The function:\n**f(x) = 1"]