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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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    $\begingroup$ Hint: if $y$ is any real number, $e^y$ is not zero. $\endgroup$
    – Chris Eagle
    Commented May 25, 2012 at 17:47
  • $\begingroup$ Be careful with logarithms! You can't just hit both sides of the equation with a logarithm, as the log of $0$ is undefined (equivalently, $b^a\neq 0$ for any real $a$ and any $b>0$)! Also, you aren't using your log rules correctly. We can't split a log over a sum or difference, and we can't pull out multiples (e.g.: $14x,21$) the way you have. $\endgroup$
    – Cameron Buie
    Commented May 25, 2012 at 18:12
  • $\begingroup$ Your first displayed line should have an “$=0$”? $\endgroup$
    – Lubin
    Commented May 25, 2012 at 18:45

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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  • $\begingroup$ Do you mean like this? $(14x - 21)/e^x^2-3x+6$ $\endgroup$
    – ninja08
    Commented May 25, 2012 at 17:53
  • $\begingroup$ Various ways of seeing it. For example, your expression is equal to $(14x-21)e^{x^3-3x+6}$. This product can only be $0$ if one of the terms is $0$. But $e^{x^2-3x+6}$ can't be $0$. So $14x-21$ must be. Or else look at the equation. For brevity let $W=e^{x^2-3x+6}$. Divide left side ($0$) by $W$. We get $0$. On the right side, divide $14xW$ by $W$. We get $14x$. Divide $21W$ by $W$. We get $21$. So $0=14x-21$. $\endgroup$
    – André Nicolas
    Commented May 25, 2012 at 18:05
  • $\begingroup$ Thank you! That explained the link I was missing. I had first plugged 0 into all the x's and assumed that $e^{x^{^{2}-3x+6}}$ would be 0 not remembering that $e^0$ = 1 $\endgroup$
    – ninja08
    Commented May 25, 2012 at 18:09
  • $\begingroup$ It is useful in many places to know what the curve $y=e^x$ looks like. $\endgroup$
    – André Nicolas
    Commented May 25, 2012 at 18:22
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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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  • $\begingroup$ If I factor everything out I'd end up where I started, which is $({14x-21})e^{x^{^{2}-3x+6}}$ $\endgroup$
    – ninja08
    Commented May 25, 2012 at 17:58

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