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# Domain

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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
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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
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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 )   Apr 18, 2022