Find the values of x for which the function f(x) = x^3 – 6x^2 – 3… - Vidyadip

Question

Find the values of $x$ for which the function $f(x) = x^3 – 6x^2 – 36x + 7$ is strictly increasing

✓

Answer

$f(x) = x^3 – 6x^2 – 36x + 7$
$\therefore f′(x) = 3x^2 – 12x – 36$
$= 3(x^2 – 4x – 12)$
$= 3(x – 6)(x + 2)$
$f(x)$ is strictly increasing, if $f′(x) > 0$
$\therefore 3(x – 6)(x + 2) > 0$
$\therefore (x – 6)(x + 2) > 0$
$ab > 0 ⇔ a > 0$ and $b > 0$ or $a < 0$ and $b < 0$
Either $x – 6 > 0$ and $x + 2 > 0$
or
$x – 6 < 0$ and $x + 2 < 0$
Case I: $x – 6 > 0$ and $x + 2 > 0$
$\therefore x > 6$ and $x > – 2$
$\therefore x > 6$
Case II: $x – 6 < 0$ and $x + 2 < 0$
$\therefore x < 6$ and $x < – 2$
$\therefore x < – 2$
Thus, $f(x)$ is strictly increasing for $x \in (−\infty −2) ∪ (6, \infty ).$

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