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Question 643090: Could you please help me with this problem? I am given the function f(x)= x^2 + 2x + 6 with a domain where x = all real numbers, and x is ≤ k. F is a one to one function, and I am to determine the greatest possible value of k. When k has that value, I am then to determine the range of f, and then the inverse function f^-1 and state its domain and range. After that I am to graph f(x) and f^-1(x). I believe I know how to find the value of k; it's just asking for the upper limit of the domain of the function given. I think to determine the range you can complete the square and put the equation into standard parabola form and so determine the vertex from that, using the y point of the vertex and the fact that it is either an up/down parabola (I believe this one is up) to find the range. I'm just having trouble finding the inverse function. When I switch the x's and y's from the original equation and try to solve for y I seem to get stuck. I know how to find the domain and range from there and I am fairly certain I can graph the equation. I just need help finding the inverse function. Please help? :)
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
In order to get the range of  you first have to establish the maximum value of  . The key to this is the fact that has been declared a 1 to 1 function. Parabolas, in their full glory, are never 1 to 1 functions because for any value of the independent variable there is a second value of the independent variable on the other side of and equidistant from the vertex that gives the same function value. In other words a full parabola never passes the horizontal line test. Enter this mysterious value . If you think about it a second, you can choose a value for at or less than the  -coordinate of the vertex and you will guarantee yourself a 1 to 1 function. Obviously, if you choose  , then will be at its maximum value.
So where is the vertex? For any quadratic function in standard form, i.e., \ =\ ax^2\ +\ bx\ +\ c) , the -coordinate of the vertex is located at  , and the  -coordinate is just the value of the function at that value.
Let's find your vertex:
\ =\ (-1)^2\ +\ 2(-1)\ +\ 6\ =\ 5)
Hence, your vertex is at )
Now we know that the maximum possible value of is -1, and, since we know the parabola opens upward because of the positive lead coefficient on , the value of the function at the vertex is a minimum, we can say that \ =\ \left{y\,\in\,\mathbb{R}\,:\,y\ \geq\ 5\right})
On to the inverse. Let's discover the inverse relation that would derive from the original function. As is normal, replace ) with
Now let's go about solving for in terms of . Put all the terms in the LHS, and every thing else in the RHS.
Complete the square on the LHS:
Factor the LHS perfect square:
^2\ =\ y\ -\ 5)
Take the root:
Finish up
And finally, swap the variables:
Almost there but we aren't quite ready to call this an inverse function yet -- because it is not a function. We have both sides of the "on its side" parabola and we only want one of the sides to match our original function. Since our original function was the left side of the parabola, the inverse has to be the bottom side of the resulting sideways parabola (because of the required symmetry of a function and its inverse about the line  ), hence:
\ =\ y\ =\ -1\ -\ \sqrt{x\ -\ 5})
Swapping the coordinates of the vertex on the original function will give you the vertex on the inverse. Your domain and range of the inverse should be easy to see from there.
John
My calculator said it, I believe it, that settles it
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! the graph below shows the original equation and its inverse.
the inverse equation is found as follows:
start with:
y = x^2 + 2x + 6
subtract 6 from both sides of the equation to get:
x^2 + 2x = y - 6
complete the squares on the left side of the equation to get:
(x+1)^2 = y - 6 + 1 which becomes:
(x+1)^2 = y - 5
take the square root of both sides of the equation to get:
x+1 = +/- sqrt(y-5)
+/- means plus or minus
subtract 1 from both sides of the equation to get:
x = -1 +/- sqrt(y-5)
replace x with y and y with x to get:
y = -1 +/- sqrt(x-5)
that's your inverse equation as seen on the graph.
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