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Question

Find the intervals on which f(x)=x4-8x2 is increasing and decreasing.


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Solution

Step 1. Find the critical point

Given function: f(x)=x4-8x2

The first derivative of the given function will be,

f'(x)=4x3-8(2)xf'(x)=4x3-16x

Now, we equate f'(x)to zero.

4x3-16x=04x(x2-4)=04x(x-2)(x+2)=0[a2-b2=(a-b)(a+b)]

Hence,the critical points are x=0,x=-2and x=2.

Step 2. Find the interval of increase and decrease

So, the four intervals formed here will be, (-,-2),(-2,0),(0,2) and (2,)

Now, we can identify the increasing and decreasing range by putting values of these intervals in f'(x).

Replace, -1-2,0

f'(-1)=4-13-16-1=-4+16=8>0

and, 32,

f'(3)=433-163=108-48=60>0

Here, we can observe that the value of f'(x) is positive in the range (-2,0)(2,).

Hence it is increasing in that range.

Now, we have -3-,-2

f'(-3)=4-33-16-3=-108+48=-60<0

and, 10,2

f'(1)=413-161=4-16=-8<0

Here, we can observe that the value of f'(x) is negative in the range (,2)(0,2).

Hence it is decreasing in that range.

Thus, the function f(x)=x4-8x2 is increasing in the interval (-2,0)(2,), and decreasing in the interval (,2)(0,2).


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