When finding the inverse of a function, for example

 

f(x) = 7-2(x-1)^2, x is greater than 1,

 

is the restriction on the inverse of this function y is greater than 1?
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Assuming you mean that y=f(x), the restrictions applied to x, the value you input or to the function (x is the domain of the function), affect the range (f(x) or y) according to what the function does. The value of y depends on x. If the inverse of the function is worked out so that x=g(y), y becomes the domain and x is the range, so restrictions on y affect x. Going back to f(x) we apply the restriction x>1 and see what affect it has on y. In this case y<7 when x>1. For g(y), the inverse restriction would be y<7. But this does not restrict x to x>1, because x can be any value. It just so happens that this f(x) function (parabola) has a maximum at f(x)=7. The easiest way to appreciate the situation is to draw the graph, which will consist of the right half of a parabola only. So the graph only exists for values of y<7 even if there were no restrictions on x. The restriction just eliminates the exact point (1,7).

g(y)=1+sqrt((7-y)/2) is the inverse, and y<7 is the restriction to match x>1. If y>7 the square root cannot be evaluated as a real quantity.

 

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