Find the critical points and the intervals of the increase and decrease of the function f(x)=(x+5)^2(x-2)^5.

veXslefsskink09

veXslefsskink09

Open question

2022-08-19

Can anyone confirm if my answer is right?
Find the critical points and the intervals of the increase and decrease of the function f ( x ) = ( x + 5 ) 2 ( x 2 ) 5
The critical points are: 3 ,   5 ,   2
The four intervals -
from left to right these open intervals are ( , 5 ) ,   ( 5 , 3 ) ,   ( 3 , 2 )   and   ( 2 , )
This part I am not sure of:
On the interval ( , 5 ), f(x) is decreasing
On the interval ( 5 , 3 ), f(x) is decreasing
On the interval ( 3 , 2 ), f(x) is increasing
On the interval ( 2 , ), f(x) is increasing

Answer & Explanation

Rayan Ali

Rayan Ali

Beginner2022-08-20Added 8 answers

Step 1
f ´ ( x ) = [ ( x + 5 ) 2 ] ´ ( x 2 ) 5 + ( x + 5 ) 2 [ ( x 2 ) 5 ] ´
f ´ ( x ) = 2 ( x + 5 ) ( x 2 ) 5 + 5 ( x + 5 ) 2 ( x 2 ) 4
f ´ ( x ) = ( x + 5 ) ( x 2 ) 4 [ 2 ( x 2 ) + 5 ( x + 5 ) ]
f ´ ( x ) = 7 ( x + 5 ) ( x + 3 ) ( x 2 ) 4
Step 2
Your break points are: 5 , 3 , 2.
Using a sign analysis you find out the intervals to be:
Increasing on: ( , 5 ) ( 3 , 2 ) ( 2 , ).
Decreasing on: ( 5 , 3 )
dammeym

dammeym

Beginner2022-08-21Added 3 answers

Step 1
First your critical points should be x = 5 , 1.5 , 2
Since f ( x ) = 4 x 3 + 18 x 2 22 x 60 and we look at the interval < x < 5 and its sign (f'(x)) is negative in that interval so it is decreasing. Next we look at 5 < x < 1.5 where the sign of the differential is positive so it is increasing.
Step 2
Then 1.5 < x < 2 where again the differential is negative so decreasing! Finally 2 < x < where the sign is positive so increasing!

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