Solution :
We have, f(x) = \(-x^2 – 2x + 15\)
\(\implies\) f'(x) = -2x – 2 = -2(x + 1)
for f(x) to be increasing, we must have
f'(x) > 0
-2(x + 1) > 0
\(\implies\) x + 1 < 0
\(\implies\) x < -1 \(\implies\) x \(\in\) \((-\infty, -1)\).
Thus f(x) is increasing on the interval \((-\infty, -1)\).
for f(x) to be decreasing, we must have
f'(x) > 0
-2(x + 1) < 0
\(\implies\) x + 1 > 0
\(\implies\) x > -1 \(\implies\) x \(\in\) \((-1, \infty)\).
Thus f(x) is decreasing on the interval \((-1, \infty)\).
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