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Some functions that aren't invertible can be made invertible by restricting their domains. For example, the function x^2 is invertible if we restrict x to the interval [0inf), or to any subset of that interval. In that case, the inverse function is sqrt(x). (We could also restrict to the domain (-inf,0], in which case the inverse function would be -sqrt(x).) Similarly, by restricting the domain of the function f(x) = 2x^2 - 4x - 9 to an interval, we can make it invertible. What is the largest such interval that includes the point x = 0?

 Oct 18, 2020
 #1
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Find the turning point (vertex) 

 

It occurs where \(\frac{-b}{2a}\) = \(\frac{-(-4)}{2(-9)}\)\(\frac{4}{-18}=\frac{2}{-9}\)

Since \(x=\) \(0\) is to the left of \(x = 1\) the larger interval is \((-\infty,\frac{2}{-9})\)

 

(Not sure if this is the correct answer, but it's what I got.)

 
 Oct 18, 2020
 #2
avatar+111124 
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Seems to make sense to me, thanks guest.  (you just made a small careless error)

 

f(x) = 2x^2 - 4x - 9

axis of symmetry   \(x=\frac{-b}{2a}=\frac{4}{4}=1\)

 

so the range would be   \((-\infty,1]\)

 
 Oct 19, 2020

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