Question 1082928
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^3\ -\ 3x^2\ +\ 5x\ =\ 2]


doesn't make any sense, so I figure you either meant +2 and you failed to hold the shift key when you pressed the +/= key, or you meant -2 and you simply pressed the +/= key instead of the _/- key.  Either scenario being equally likely, I'm going to do the problem for the following function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ x^3\ -\ 3x^2\ +\ 5x\ +\ a]


Where *[tex \Large a] is an arbitrary real number constant.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{dy}{dx}\ =\ 3x^2\ -\ 6x\ +\ 5]


The minimum slope tangent to the original function will occur at the point where the first derivative has a minimum.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{d^2y}{dx^2}\ =\ 6x\ -\ 6]


Which has a zero at *[tex \Large x\ =\ 1]


For the original function


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(1)\ =\ 3\ +\ a]


Hence, the point of tangency is *[tex \Large (1,\,3\,+\,a)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f'(1)\ =\ 2]


So you want an equation of a line that passes through the point *[tex \Large (1,\,3\,+\,a)] with a slope of 2.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ (3\ +\ a)\ =\ 2(x\ -\ 1)]


Once you figure out whether *[tex \Large a\ =\ 2] or *[tex \Large a\ =\ -2],


Then you can complete the problem.


By the way, the question was also worded incorrectly in the first place.  You cannot find THE equation of a line.  You can only find AN equation of a line.
  


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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