5) Find the intervals in which the function f given by $f (x) = 2x^3 - 3x ^2 - 36 x + 7$ is
a) increasing

Answers (1)

G Gautam harsolia

It is given that
$f (x) = 2x^3 - 3x ^2 - 36 x + 7$
So,
$f^{'}(x)= 6x^{2} - 6x - 36$
$f^{'}(x)= 0$
$6x^{2} - 6x -36 =0 \Rightarrow 6 (x^{2} - x-6)$
$x^{2} - x-6 = 0$
$x^{2} - 3x+2x-6 = 0$
$x(x-3) + 2(x-3) = 0\\$
$(x+2)(x-3) = 0$
x = -2 , x = 3

So, three ranges are there $(-\infty,-2) , (-2,3) \ and \ (3,\infty)$
Function $f^{'}(x)= 6x^{2} - 6x - 36$ is positive in interval $(-\infty,-2) , (3,\infty)$ and negative in the interval (-2,3)
Hence, $f (x) = 2x^3 - 3x ^2 - 36 x + 7$ is strictly increasing in $(-\infty,-2) \cup (3,\infty)$
and strictly decreasing in the interval (-2,3)