Question 1055355
Converting to vertex form,
{{{f(x)=2(x-0)^2+2}}}
So the domain is all real numbers, the range is [{{{2}}},{{{infinity}}})
To get the inverse, use x,y nomenclature,
{{{y=2x^2+2}}}
Interchange x and y and solve for y.
The new y is the inverse.
{{{x=2y^2+2}}}
{{{2^y^2=x-2}}}
{{{y^2=(x-2)/2}}}
{{{y=0 +- sqrt((x-2)/2)}}}
{{{f^(-1)=0 +- sqrt((x-2)/2)}}}
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