Find the interval in which the following function is strictly incerasing or decreasing,
6−9x−x2
Given,f(x)=6−9x−x2⇒f′(x)=−9−2x
On putting, f(x)=6−9x−x2⇒f′(x)=−9−2x
On putting f'(x)=0, we get −9−2x=0⇒x=−92
Which divides the real line in two disjoint
intervals (−∞,−92) and (−92,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−92)+veStrictly increasing(−92,∞)−veStrictly decreasing
Therefore, f(x) is strictly increasing when x<−92 and strictly decreasing when x>−92.