Q. 205.0( 1 Vote )

# Find the interval

Given Function:

f(x)= (x–1)3(x+2)2

To find points where function touches the x axis are

f'(x)=0

f'(x)=3(x–1)2(x+2)2 + 2(x–1)3(x+2)=0

(x–1)2(x+2)(3x+6+2x–2)=0

(x–1)2(x+2)(5x+4)=0

Since the function is defined for all real x

Therefore it divides it in 4 intervals:

We will check the sign of for every interval if sign is positive then function is increasing vice–versa

f'(x) = (x–1)2(x+2)(5x+4)

1. In Interval (–∞,–2),

• Sign of f’(x) is positive. Therefore f(x) is increasing in this interval.

2.

• Sign of f’(x) is negative. Therefore f(x) is decreasing in this interval.

3.

• Sign of f’(x) is positive. Therefore f(x) is increasing in this interval.

4. In Interval (1,∞)

• Sign of is positive. Therefore is increasing in this interval.

For finding the local minimum and local maximum, we have to consider first derivative test.

If left hand derivative at point is positive and right hand derivative is negative then the point is local maximum.

If left hand derivative at point is negative and right hand derivative is positive then the point is local minimum.

Minima and Maxima always has to search where derivative changes the sign.

Using above information

So –2 is a point of local maximum

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