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1 answer:

4 0

Por determinante:
5...1...1
1...0...1 = 0 --> se o determinante for igual a 0, os pontos estarão alinhados.
3...3...1

calculando: (5 x 0 x 1 + 1 x 1 x 3 + 1 x 3 x 1) - (1 x 0 x 3 + 1 x 1 x 1 + 3 x 1 x 5) =
0 + 3 + 3 - 0 - 1 - 15 =
6 - 1 - 15 =
5 - 15 = - 10 --> como não deu zero, os pontos não estão alinhados.

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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .