algebra-precalculuscalculus
Determine the values of x where the tangent line is horizontal for the function:$$\frac {1}{(x^2-16)(x-7)}$$
The value(s) of $x$ where the tangent line to the graph of the function is horizontal is(are)_.
The Derivative of the $f(x)$ is
$$f'(x)=\frac{-3x^2+14x-16}{(x-7)^2(x^2-16)^2}$$
Apply the quotient rule to find $$ y' = \frac{e^x \cos x - e^x \sin x}{e^{2x}} $$
We want $y'=0$, which means we want the numerator to equal zero:
$$ 0 = e^x \cos x - e^x \sin x $$ $$ 0 = \cos x - \sin x $$ $$ \sin x = \cos x $$ Divide by $\cos x$: $$ \tan x = 1 $$
This happens at $x = \frac{\pi}{4}$, and since tangent has period $\pi$, it will happen also when we add an integer multiple of $\pi$. Thus, we have a horizontal tangent line at $$ x = \frac{\pi}{4} + n\pi, n \in \mathbb{Z} $$
If you want a horizontal tangent, then you need to look for where ${dy\over dx}=0$.
Edit: As indicated, solve ${dy\over dx}=0\implies y(2.5-1/x)=0\implies y=0\text{ or }x={2\over 5}$. But $y=0$ is not in the domain of the original relation, $\ln(xy)=2.5x$, so the horizontal tangent occurs only when $x=2/5\implies y=5e/2$.
This is depicted graphically below: the dashed line is the horizontal tangent passing thru $(2/5,5e/2)$ on the graph of $\ln(xy)=2.5x$.
Best Answer
You are almost on the right track (your numerator should be $-3x^2+14x+16$). Once you've fixed that, you'll want to figure out when the numerator of the derivative is $0$.
A good thing to keep in mind when dealing with rational functions is this: for a tangent line to exist at a point (horizontal or otherwise), the function must be defined at that point. So, we'll need to toss out solutions to $f'(x)=0$ that are not in the domain of $f(x)$ (if there are any such solutions).