Question 263527
we're supposed to find the inverse of the power function, my teacher is lazy and dosnt teach us much. the answer i got was {{{ +- 2 sqrt one halfx-1}}} he marked it wrong...
<pre><font size = 4 color = "indigo"><b>
First of all, the answer you gave is not a function, with the
"positive or negative".  No function can have a "±" in it.

Let's draw the graph of f(x) first.

Now if the domain {{{x<=0}}} weren't given and we just plotted

{{{f(x)=2-2x^2}}}

we'd have this graph:

{{{graph(400,400,-5,5,-5,5,2-2x^2)}}}

That graph passes the vertical line test, because no vertical
line crosses the graph more than once, so it is a function.
However it is not a one-to-one function because it does not
pass the horizontal line test.  If you draw a horizontal line
through it it will cut the graph twice.  So it does not have
an inverse function.

However since {{{x<=0}}} is given for its domain, the graph
of f(x) is NOT the one above.  Here is the graph of

{{{f(x)=2-2x^2}}} with the given domain {{{x<=0}}}

{{{graph(400,400,-5,5,-5,5,(2-2x^2)*sqrt(-x)/sqrt(-x))}}} 

It consists only of the left half of the graph of {{{f(x)=2-2x^2}}}

without the given domain {{{x<=0}}}

So you see now that the function passes not only the VERTICAL line
test, it passes the HORIZONTAL line test as well.  So with that
restricted domain, it DOES have an inverse.  

Now let's sketch the inverse before finding its equation.  First draw 
the line whose equation is {{{y=x}}}, called the IDENTITY line 
(because x and y are indentically equal if you have y=x).  Let's 
draw that line.  It is a line through the origin which goes at a 
45° angle to both axes.  I'll draw it dotted and in green

{{{drawing(400,400,-5,5,-5,5, 

graph(400,400,-5,5,-5,5,-6,x*sqrt(sin(10x))/sqrt(sin(10x))),

graph(400,400,-5,5,-5,5,(2-2x^2)*sqrt(-x)/sqrt(-x)))}}} 

Now the inverse of f(x) is the reflection of the graph of

f(x) into that green dotted identity line, whose equation is y=x.

I'll draw it in blue:

{{{drawing(400,400,-5,5,-5,5, 

graph(400,400,-5,5,-5,5,-6,x*sqrt(sin(10x))/sqrt(sin(10x)),-sqrt((2-x)/2)),

graph(400,400,-5,5,-5,5,(2-2x^2)*sqrt(-x)/sqrt(-x)))}}} 

The inverse of a function is a function symmetric with the
original function's graph about the identity line, y=x.

Now let's find the equation of that inverse.

{{{f(x)=2-2x^2}}}, x<= 0

1. Change f(x) to y

{{{y=2-2x^2}}}

2. Interchange x and y

{{{x=2-2y^2}}}

3.  Solve for y:

{{{2y^2=2-x}}}

{{{y^2=(2-x)/2}}}

{{{y=""+-sqrt((2-x)/2)}}}

Now looking at the graph, the blue inverse graph
is entirely below the x-axis.  Therefore
we can only take the negative square root, 
and ignore the positive square root.

{{{y=-sqrt((2-x)/2)}}}

4.  Finally change y to {{{f^(-1)}}}{{{"(x)"}}}

So the final answer is

{{{f^(-1)}}}{{{"(x)"}}}{{{""=""}}}{{{-sqrt((2-x)/2)}}}

Your teacher might expect you to rationalize the denominator.

If so the final answer is

{{{f^(-1)}}}{{{"(x)"}}}{{{""=""}}}{{{-sqrt(2(2-x))/2}}}

or

{{{f^(-1)}}}{{{"(x)"}}}{{{""=""}}}{{{(-sqrt(4-2x))/2}}}

Edwin</pre>