The maximum value of the function \(f(x) = 2^{3 + 4x - 2x^{2}}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 32
Differentiating, with \(\frac{d}{dx}2^{u} = 2^{u}\log_e 2 \cdot \frac{du}{dx}\):
\(f'(x) = 2^{3 + 4x - 2x^{2}} \cdot \log_e 2 \cdot (4 - 4x)\)
The factors \(2^{3 + 4x - 2x^{2}}\) and \(\log_e 2\) are positive, so f′(x) has the sign of 4 − 4x:
\(f'(x) \gt 0\) for \(x \lt 1\)
\(f'(x) \lt 0\) for \(x \gt 1\)
By the first derivative test, f has its maximum at x = 1:
\(f(1) = 2^{3 + 4 - 2} = 2^{5} = 32\)