f:R-->R. f(x)=-2x^2+12x-13
What is Image of f(x)?
second question:
Divide the domain R to 2 intervals D1 and D2 to get two inverse functions in these intervals. Calculate these functions.
The image is R, because f hasnīt restriction, you can draw the grafh and you will see that Im (f) = R
To find the inverse you must divide R in two intervalo, you must find de maximun or minimun, we say that this point is X, then:
$\displaystyle
D_1 = \left( { - \infty ,x} \right) \wedge D_2 = \left( {x,\infty } \right)
$
I'm afraid Nacho is incorrect. He may be thinking of the "domain" rather than the "range" or image. Complete the square on that quadratic: $\displaystyle f(x)=-2x^2+12x-13= -2(x^2+ 6x+ 9)- 13+ 18= -2(x-3)^2+ 6$. Now it should be clear that f(x) is never larger than 6. And that should make it clear how to solve for x to find the two "inverse" functions.
Ohh.. now it's a little bit clearer...
Some questions:
-How can I prove that f:R-->(inf,6] is not inverse?
-What are the two intervals that I must take to prove that f is inverse.
I mean: f_1-->(inf,6] and f_2-->(inf,6], which intervals should I take (D_1, D_2)?
And thank you a lot!
I'm sorry, HallsofIvy undertands my little confunsion
My teacher always say that he doesnīt like this method, but it works fine
$\displaystyle
\begin{gathered}
y = - 2x^2 + 12x - 13 \Leftrightarrow - 2x^2 + 12x - 13 + y = 0 \hfill \\
\Rightarrow x = \frac{{ - 12 \pm \sqrt {12^2 - 4 \cdot - 2 \cdot \left( { - 13 + y} \right)} }}
{{2 \cdot - 2}} = \frac{{ - 12 \pm \sqrt {144 + 8 \cdot \left( {y - 13} \right)} }}
{{ - 2}} \hfill \\
\end{gathered}
$
If $\displaystyle
x \in \left( { - \infty ,6} \right) \Rightarrow x = \frac{{ - 12 - \sqrt {144 + 8 \cdot \left( {y - 13} \right)} }}
{{ - 2}}
$
Now you changes the X for the Y
Then your inverse is: $\displaystyle
f^{ - 1} :\left( { - \infty ,6} \right) \to \mathbb{R},{\text{ }}f^{ - 1} \left( x \right) = \frac{{ - 12 - \sqrt {144 + 8 \cdot \left( {x - 13} \right)} }}
{{ - 2}}
$
To $\displaystyle
x \in \left( {6,\infty } \right)
$ is analogous