Question 1189255
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ +\ kx\ +\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{df}{dx}\ =\ 2x\ +\ k]


*[tex \Large f(x)] is a quadratic polynomial with a positive lead coefficient so it has a minimum value where *[tex \Large \frac{df}{dx}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ k\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{min}\ =\ \frac{-k}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f\(x_{min}\)\ =\ \(\frac{-k}{2}\)^2\ +\ 4\(\frac{-k}{2}\)\ +\ 8]

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ 1\ +\ 4x\ -\ 2x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dg}{dx}\ =\ 4\ -\ 4x]


*[tex \Large g(x)] is a quadratic polynomial with a negative lead coefficient so it has a maximum value where *[tex \Large \frac{dg}{dx}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ 4x\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{max}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g\(x_{max}\)\ =\ 1\ +\ 4(1)\ -\ 2(1)^2\ =\ 3]


So the problem is to solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(\frac{-k}{2}\)^2\ +\ 4\(\frac{-k}{2}\)\ +\ 8\ =\ 3]


for *[tex \Large k]


You can do the necessary arithmetic.

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

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