Question 1073969
We sometimes use symbols that are difficult to interpret, or make mistakes while typing (or while thinking), but we try.
 
The function {{{f(x) = 2x^2-4x-5}}} , with all real numbers as its domain, {{{graph(300,300,-4,6,-10,10,2x^2-4x-5)}}} is not invertible,
because (like all quadratic functions) it has a vertical axis of symmetry.
That axis of symmetry is the line {{{x=1}}} : {{{drawing(300,300,-4,6,-10,10,
graph(300,300,-4,6,-10,10,2x^2-4x-5),line(1,-10,1,10)
)}}}.

By restricting the domain of the function
to a domain that does not include points on both sides of that axis,
we can make {{{f(x) = 2x^2-4x-5}}}  invertible.
Defining the function only for a set of points with {{{x>=1}}} ,
or to a set of points with {{{x<=1}}} , makes the function invertible.
If we want to include the point {{{x=0}}} in the domain,
we cannot include any point with {{{x>1}}} ,
but we can include any point with {{{x<=1}}} .
The largest restricted domain that includes the point {{{x=0}}} ,
but does not contain any point with {{{x>1}}} is
all real {{{x}}} such that {{{x<=1}}} .
That can be expressed in interval notation as
{{{"("}}}{{{-infinity}}}{{{", 1 ]"}}} or {{{highlight("( -infinity , 1 ]")}}}.
That is the largest interval that includes the point {{{x=0}}},
where we can make {{{f(x) = 2x^2-4x-5}}} invertible.
That is the largest such interval that includes the point {{{x=0}}}.
 
The equation for the axis of symmetry can be found by transforming the function, or by applying a formula.
With formulas:
All quadratic functions can be written in the form {{{y=ax^2+bx+c}}} ,
for some constants {{{a<>0}}} , {{{b}}} and {{{c}}} .
The axis of symmetry for such a function is the line {{{x=(-b)/"2 a"}}}
{{{f(x) = 2x^2-4x-5}}} is a quadratic function with {{{a=2}}} and {{{b=-4}}} ,
so its axis of symmetry is
{{{x=(-(-4))/(2*2)}}} --> {{{x=4/4}}} ---> {{{x=1}}} .
Transforming the function:
{{{f(x) = 2x^2-4x-5}}} or {{{y = 2x^2-4x-5}}}
{{{y=2x^2-4x+2-7}}}
{{{y=2(x^2-2x+1)-7}}}
{{{y=2(x-1)^2-7}}}
 
The inverse for {{{y=2(x-1)^2-7}}} {{{for}}}{{{all}}} {{{x<=1}}} ,{{{graph(300,300,-4,6,-10,10,2x^2-4x-5+sqrt(1-x)-sqrt(1-x))}}} is
{{{system(x=2(y-1)^2-7,"with",y<=1)}}} ---> {{{system((x+7)/2=(y-1)^2,"with",y-1<=0)}}} ---> {{{y-1=-sqrt((x+7)/2)}}} ---> {{{y=1-sqrt((x+7)/2)}}} ,{{{graph(300,300,-10,10,-4,6,1-sqrt((x+7)/2))}}}
or {{{f^(-1)}}}{{{(x)=1-sqrt((x+7)/2)}}} .
 
Here are the graphs for {{{blue(f(x))}}}, {{{red(f^(-1))}}}, and {{{green(y=x)}}} , which is the line you flip {{{bule(f)}}} over to get {{{red(f^(-1))}}} :
{{{graph(300,300,-10,10,-10,10,1-sqrt((x+7)/2),x,2x^2-4x-5+sqrt(1-x)-sqrt(1-x))}}}