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Let f(x) = (x+2)^2-5. If the domain of f is all real numbers, then f does not have an inverse function, but if we restrict the domain of f to an interval\( [c,\infty)\), then f may have an inverse function. What is the smallest value of c we can use here, so that f does have an inverse function?

 Apr 12, 2019
 #1
avatar+102921 
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The vertex of  f(x)  = (-2, - 5)

So...the smallest "c" we can have is -2

 

We can find the inverse as

y = ( x + 2) ^2 - 5

y + 5 = (x + 2)^2

√ [ y + 5]  = x + 2

√[y + 5 ] - 2 = x

√[x + 5 ] - 2 = y  = the inverse

 

Note that the point  (-5, -2)  is on the inverse....as we would expect

 

See the graph , here :  https://www.desmos.com/calculator/w4bhcvpkkb

 

Note that if f(x) included any x values in the domain < -2, it would not be one-to-one...and we wouldn't have a true inverse

 

cool cool cool

 Apr 12, 2019

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