1. ## Horizontal Tangent Lines

f(x)= xe^(x)/(9x+b)

How would you find the coordinates of f where the tangent lines are horizonal when b=-7?

2. Originally Posted by ctran
f(x)= xe^(x)/(9x+b)

How would you find the coordinates of f where the tangent lines are horizonal when b=-7?
If $\displaystyle b = -7$, then

$\displaystyle f(x) = \frac{xe^x}{9x - 7}$.

Tangent lines are horizontal when the derivative is $\displaystyle 0$.

So $\displaystyle f'(x) = \frac{(9x - 7)\,\frac{d}{dx}(xe^x) - xe^x\,\frac{d}{dx}(9x - 7)}{(9x - 7)^2}$

$\displaystyle = \frac{(9x - 7)(xe^x + e^x) - 9xe^x}{(9x - 7)^2}$

Now let $\displaystyle f'(x) = 0$ and solve for $\displaystyle x$.

$\displaystyle \frac{(9x - 7)(xe^x + e^x) - 9xe^x}{(9x - 7)^2} = 0$

$\displaystyle e^x(9x - 7)(x + 1) - 9xe^x = 0$

$\displaystyle e^x[(9x - 7)(x + 1) - 9x] = 0$

Since $\displaystyle e^x \neq 0$ for any $\displaystyle x$, we have

$\displaystyle (9x - 7)(x + 1) - 9x = 0$

$\displaystyle 9x^2 +9x - 7x - 7 - 9x = 0$

$\displaystyle 9x^2 - 7x - 7 = 0$

$\displaystyle x = \frac{7 \pm \sqrt{301}}{18}$.

Now, substitute these values of $\displaystyle x$ back into $\displaystyle f(x)$.