Prepare-se para um confronto eletrizante! Nesta segunda-feira, 7 de abril de 2025, às 20h30 (horário de Brasília), o Kaseya Center, em Miami, será palco do duelo entre o Miami Heat e o Philadelphia 76ers. A partida, válida pela reta final da temporada regular da NBA, promete fortes emoções, com o Heat buscando consolidar sua vaga no play-in e os 76ers tentandoFind the inverse of $f(x) = x^2 – 6x + 5$ where $x > 3$.
To find the inverse of the function $f(x) = x^2 – 6x + 5$ where $x > 3$, we first rewrite the function as $y = x^2 – 6x + 5$. We complete the square to express the equation in terms of $(x-3)^2$:
$$y = (x^2 – 6x + 9) – 9 + 5 = (x-3)^2 – 4.$$
Now we solve for $x$ in terms of $y$.
$$y = (x-3)^2 – 4 \Rightarrow y+4 = (x-3)^2$$
Taking the square root of both sides, we get
$$\sqrt{y+4} = |x-3|.$$
Since $x > 3$, we have $x-3 > 0$, so $|x-3| = x-3$. Then
$$x-3 = \sqrt{y+4} \Rightarrow x = \sqrt{y+4} + 3.$$
Now we switch $x$ and $y$ to find the inverse function:
$$y = \sqrt{x+4} + 3.$$
The inverse function is $f^{-1}(x) = \sqrt{x+4} + 3$.
Since $x > 3$, we have $x-3 > 0$, so $x-3 = \sqrt{y+4}$ and $(x-3)^2 = y+4$, which means $y = (x-3)^2 – 4 > -4$. So the domain of the inverse function is $x > -4$.
We can check that $f^{-1}(f(x)) = x$ for $x > 3$:
\begin{align*} f^{-1}(f(x)) &= f^{-1}(x^2-6x+5) = f^{-1}((x-3)^2 – 4) \\ &= \sqrt{(x-3)^2 – 4 + 4} + 3 = \sqrt{(x-3)^2} + 3 \\ &= |x-3| + 3. \end{align*}
Since $x > 3$, $|x-3| = x-3$. Thus, $f^{-1}(f(x)) = x-3+3 = x$.
Final Answer: The final answer is $\boxed{f^{-1}(x) = \sqrt{x+4}+3}$
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