The given equation is:
\(2^x + 2^{-x} = 2 - (x - 2)^2\)
Analysis of each side follows.
Left-Hand Side (LHS): \(2^x + 2^{-x}\)
- For any real \(x\), both \(2^x\) and \(2^{-x}\) are positive.
- By the AM ≥ GM inequality:
\[ 2^x + 2^{-x} \geq 2\sqrt{2^x \cdot 2^{-x}} = 2 \] Equality holds when \(2^x = 2^{-x}\), which implies \(x = 0\). - Therefore, LHS ≥ 2, with equality occurring exclusively at \(x = 0\).
Right-Hand Side (RHS): \(2 - (x - 2)^2\)
- For any real \(x\), \((x - 2)^2\) is non-negative.
- The maximum value of the RHS is achieved when \((x - 2)^2 = 0\), which occurs at \(x = 2\).
- At \(x = 2\), RHS = 2. For all other values of \(x\), RHS<2.
Comparison of LHS and RHS
- The LHS is always ≥ 2, reaching equality at \(x = 0\).
- The RHS is always ≤ 2, reaching equality at \(x = 2\).
- Consequently, no value of \(x\) exists for which both sides are equal.
Thus, there are no real solutions to the equation.
Final Answer: \(0\) real-valued solutions.