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The domain of the function q(x) = x^4 +4x^2 + 4 - 12x^2 - 18  is [0,inf). What is the range?

 Apr 17, 2022
 #1
avatar+9369 
+1

Note that:

 

\(\begin{array}{rcl} q(x) &=& x^4 + 4x^2 + 4 - 12x^2 - 18\\ &=& (x^4 + 4x^2 +4) - 12x^2 - 24 + 6\\ &=& (x^4 + 4x^2 +4) - 12(x^2 + 2) + 6\\ &=& (x^2 + 2)^2 -12(x^2 + 2) + 6\\ &=& (x^2 + 2)^2 - 12(x^2 + 2) + 36 - 30\\ &=& ((x^2 + 2)^2 - 12(x^2 + 2) + 36) - 30\\ &=& (x^2 - 4)^2 - 30 \end{array}\)

The minimum of q(x) is attained at x = 2, which is \(q(2) = (2^2 - 4)^2 - 30 = -30\).

Therefore, the range of q(x) in [0, inf) is \([-30, \infty)\).

 Apr 17, 2022
 #2
avatar+122390 
+1

Because of the   leading x^4 term,  the max  is unbounded    =    inf

 

Using some Calculus  to find the min

 

q'(x)  = 4x^3 + 8x - 24x

 

q'(x)  =  4x^3 - 16x

 

Set this  =  0

 

4x^3 -16x   = 0

 

4x (x^2 - 4)  = 0 

 

x=  0     or  x   =2

 

Taking the second derivative

 

12x - 16

 

When x = 0     this is negative .....so....not a min here

 

When x =  2  this is positive, so  x is a min here

 

Petting 2 back into  the  function and evaluating  gives us    -30   

 

So

 

Range  =   [ -30 , inf )

 

cool cool cool

 Apr 18, 2022

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