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Home » \fracA_\textcircleA_\texttriangle = \frac\pi r^2r s = \frac\pi rs - Veritas Home Health

\fracA_\textcircleA_\texttriangle = \frac\pi r^2r s = \frac\pi rs - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Solution: Start by using the identity $ \sin^2 x = 1 - \cos^2 x $ to r...
f(x) = 1 - \cos^2 x + 2\cos x = -\cos^2 x + 2\cos x + 1.
Let $ y = \cos x $, where $ y \in [-1, 1] $. The function becomes:
f(y) = -y^2 + 2y + 1.
This is a quadratic opening downward. Its maximum occurs at the vertex...
f(1) = -1 + 2 + 1 = 2.
Evaluate at $ y = -1 $:
f(-1) = -1 - 2 + 1 = -2.
Thus, the range of $ f(x) $ is $[-2, 2]$.
\boxed{[-2, 2]}

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