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I started here:

$({14x-21})e^{x^{^{2}-3x+6}}$

took the derivative to end up here:

$0=14xe^{x^{2}-3x+6}-21e^{x^{2}-3x+6}$

and now I must solve for x. Or, find where the tangent line's slope is horizontal.

I'm stuck here, though I'm not sure if this was the right procedure.

$14x(x^2-3x+6)(\ln e) - 21(x^2-3x+6)(\ln e)$

Edit:

The answer is $3/2$ I just don't know how to get there.

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    Your first displayed line should have an “$=0$”?2012-05-25

2 Answers 2

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Divide everything by $e^{x^2-3x+6}$. It's OK, the thing can't be $0$.

The taking of logarithms process was dead wrong. In general, $\ln(x-y)\ne \ln x-\ln y$. And that wasn't the only mistake made in finding the logarithm. For example, note that $\ln(14xe^{x^{2}-3x+6})=\ln 14 +\ln x +x^2-3x+6$.

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    It is useful in many places to know what the curve $y=e^x$ looks like.2012-05-25
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Factor out $14x-21$ then solve the equation. The exponential function is always non-zero, unless the exponent is minus infinity, which is not possible for your exponent. So the only solution is $x = 21/14$.

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    If I factor everything out I'd end up where I started, which is $({14x-21})e^{x^{^{2}-3x+6}}$2012-05-25