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Given #f'(x) = (x+1)(x-2)²g(x)# where #g# is a continuous function and #g(x) < 0# for all #x#. On what interval(s) is #f# decreasing?

Answer 1

Scarlett Allison

#x > -1#

For #f(x)# to be decreasing is necessary that

#f'(x) < 0# so

#(x+1)(x-2)^2g(x) < 0#

If #g(x) < 0 forall x in RR# then this is equivalent to

#(x+1)(x-2)^2 > 0#

Here #(x-2)^2 ge 0# so the condition is

#x+1 > 0# or #x > -1#

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Answer 2

Scarlett Allison

To determine the intervals on which ( f(x) ) is decreasing, we need to analyze the sign of ( f'(x) ).

Given ( f'(x) = (x+1)(x-2)^2 g(x) ), and knowing that ( g(x) < 0 ) for all ( x ), we can focus on the signs of ( (x+1) ) and ( (x-2)^2 ).

( (x+1) ) is positive for ( x > -1 ) and negative for ( x < -1 ).
( (x-2)^2 ) is always positive for real ( x ).

Now, when ( g(x) < 0 ), and both factors in ( f'(x) ) are negative, ( f'(x) ) is positive, which indicates that ( f(x) ) is increasing.

Thus, the intervals where ( f(x) ) is decreasing are where ( (x+1) ) is negative and ( (x-2)^2 ) is positive.

This occurs for ( -1 < x < 2 ). Therefore, ( f(x) ) is decreasing on the interval ( -1 < x < 2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Given #f'(x) = (x+1)(x-2)²g(x)# where #g# is a continuous function and #g(x) &lt; 0# for all #x#. On what interval(s) is #f# decreasing?

Given #f'(x) = (x+1)(x-2)²g(x)# where #g# is a continuous function and #g(x) < 0# for all #x#. On what interval(s) is #f# decreasing?