# 6) Find the intervals in which the following functions are strictly increasing or decreasing:e) $( x+1) ^3 ( x-3) ^3$

G Gautam harsolia

Given function is,
$f(x) = ( x+1) ^3 ( x-3) ^3$
$f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$
Now,
$f^{'}(x) = 0 \\ 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^{3} \\ 3(x+1)^{2}(x-3)^{2}((x-3) + (x+1) ) = 0 \\ (x+1)(x-3) = 0 \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ (2x-2) = 0\\ x=-1 \ and \ x = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ x = 1$
So, the intervals are  $(-\infty,-1), (-1,1), (1,3) \ and \ (3,\infty)$

Our function $f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$   is +ve in the interval  $(1,3) \ and \ (3,\infty)$
Hence,  $f(x) = ( x+1) ^3 ( x-3) ^3$  is strictly increasing in the interval  $(1,3) \ and \ (3,\infty)$
Our function $f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$   is -ve in the interval  $(-\infty,-1) \ and \ (-1,1)$
Hence,  $f(x) = ( x+1) ^3 ( x-3) ^3$  is strictly decreasing in interval  $(-\infty,-1) \ and \ (-1,1)$

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