Q

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

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

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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)

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