When is the tangent line of the following function horizontal?
$$y =\frac{\sin x}{e^x}$$
What steps or how can I draw a graph to figure this out or prove this?
[Math] When is tangent line horizontal
calculus
Related Solutions
Simplify $f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$
- $f'(x) = (2x^2 -5x + 2) + (x^2 - x - 11)$
- $f'(x) = 3x^2 - 6x - 9$
And find where $f'(x) = 0$
$3x^2 - 6x - 9 = 0 \iff x^2 - 2x - 3 = (x + 1)(x - 3) = 0$
That's where slope is 0, hence any line tangent at that point will be horizontal: when $x = 3$ or when $x = -1$.
So the roots (x values) of the points you need are
$x_1 = 3$, and
$x_2 = -1$.
Then find the corresponding $y$ value in the ORIGINAL equation: $$\;f(x)=(x-2)(x^2-x-11),\;$$ to determine the two points: $(x_1, y_1), (x_2, y_2)\;$ where the line tangent to $f(x)$ is horizontal.
$y_1 = f(3) = 1 \cdot -5 = -5$
$y_2 = f(-1) = (-3)(-9) = 27$
So your points are $(3, -5)$ and $(-1, 27)$.
I've included a graph of the function $\;f(x)=(x-2)(x^2-x-11)\;$(in blue), along with the two horizontal lines tangent to the function: $\;y = -5,\;\; y = 27$ (violet and "brown", respectively) to see in a "picture" what is happening here.
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).
Best Answer
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 $$ x = \frac{\pi}{4} + n\pi, n \in \mathbb{Z} $$