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# function

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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
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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