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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 b2a = (4)2(9)418=29

Since x= 0 is to the left of x=1 the larger interval is (,29)

 

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

 Oct 18, 2020
 #2
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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=b2a=44=1

 

so the range would be   (,1]

 Oct 19, 2020

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