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  • Find the intervals in which the following functions are strictly increasing or decreasing x plus 1 the whole cube x minus 3 the whole cube

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

Answers (1)

best_answer

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)

Posted by

Gautam harsolia

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