Question 521599
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ +\ 2x\ +\ 2\ \Rightarrow\ y\ =\ (x\ +\ 1)^2\ +\ 1\ \Rightarrow\ y\ \geq\ 1]


Since


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


is a concave up parabola such that the vertex is a minimum point and where the *[tex \Large x]-coordinate of the vertex is equal to *[tex \Large \frac{-2}{2}\ =\ -1] and the *[tex \Large y]-coordinate of the vertex is *[tex \Large (-1)^2\ +\ 2(-1)\ +\ 2\ =\ 1], the minimum value of the function is, indeed, 1 and the given proposition is true.


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