Question 643090
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In order to get the range of *[tex \LARGE f] you first have to establish the maximum value of *[tex \LARGE k].  The key to this is the fact that *[tex \LARGE f] 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 *[tex \LARGE k].  If you think about it a second, you can choose a value for *[tex \LARGE k] at or less than the *[tex \LARGE x]-coordinate of the vertex and you will guarantee yourself a 1 to 1 function.  Obviously, if you choose *[tex \LARGE k\ =\ x_v], then *[tex \LARGE k] will be at its maximum value.


So where is the vertex?  For any quadratic function in standard form, i.e., *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c], the *[tex \LARGE x]-coordinate of the vertex is located at *[tex \LARGE x_v\ =\ \frac{-b}{2a}], and the *[tex \LARGE y]-coordinate is just the value of the function at that value.


Let's find your vertex:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ \frac{-2}{2}\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v\ =\ f(x_v)\ =\ (-1)^2\ +\ 2(-1)\ +\ 6\ =\ 5]


Hence, your vertex is at *[tex \LARGE (-1,5)]


Now we know that the maximum possible value of *[tex \LARGE k] is -1, and, since we know the parabola opens upward because of the positive lead coefficient on *[tex \LARGE f], the value of the function at the vertex is a minimum, we can say that *[tex \LARGE \text{ran}(f)\ =\ \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 *[tex \LARGE f(x)] with *[tex \LARGE y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ +\ 2x\ +\ 6]


Now let's go about solving for *[tex \LARGE x] in terms of *[tex \LARGE y].  Put all the *[tex \LARGE x] terms in the LHS, and every thing else in the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ =\ y\ -\ 6]


Complete the square on the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ y\ -\ 5]


Factor the LHS perfect square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)^2\ =\ y\ -\ 5]


Take the root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ \pm\sqrt{y\ -\ 5}]


Finish up


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1\ \pm\sqrt{y\ -\ 5}]


And finally, swap the variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ y\ =\ -1\ \pm\sqrt{x\ -\ 5}]


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 *[tex \LARGE y\ =\ x]), hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ 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
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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