Find the set of all real values of the parameter \(m\) for which the function \(y = – {x^3} – 6{x^2} + \left( {4m – 9} \right)x + 4\) is strictly decreasing over the interval \(\left( { – \infty ; – 1} \right)\) as
A. \(\left( { – \infty ; – \frac{3}{4}} \right]\)
B. \(\left[ {0; + \infty } \right)\)
C. \(\left( { – \infty ;0} \right]\)
D. \(\left[ { – \frac{3}{4}; + \infty } \right)\)
Solution
We have \(y’ = – 3{x^2} – 12x + 4m – 9\)
To have the function strictly decreasing over the interval \(\left( { – \infty ; – 1} \right)\), we need \(y’ \le 0\) for all \(x \in \left( { – \infty ; – 1} \right)\).
This implies \(3{x^2} + 6x + 9 \ge 4m\) for all \(x \in \left( { – \infty ; – 1} \right)\).
Let \(f\left( x \right) = 3{x^2} + 12x + 9\).
We have \(f’\left( x \right) = 6x + 12;\,\)\(f’\left( x \right) = 0 \Leftrightarrow x = – 2\).
We find that \(f\left( x \right) \ge 4m\) for all \(x \in \left( { – \infty ;1} \right)\) if and only if \(4m \le – 3 \Leftrightarrow m \le \frac{{ – 3}}{4}\).
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