Find the point(s) on the graph of f where the tangent line is horizontal. $f(x)=x^{8} e^{-x}$
A) $\left(8, \frac{8^{8}}{e^{8}}\right)$
B) (0, 0)
C) (0, 0), $\left(8, \frac{8^{8}}{e^{8}}\right)$
D) $\left(1, \frac{1}{e}\right)$

Question

Quizplus · Accepted Answer

To find where the tangent line is horizontal, we need to find where the derivative $f'(x)$ is equal to zero. We can use the product rule and chain rule to find the derivative: 
$$f(x) = x^8e^{-x}$$
$$f'(x) = 8x^7e^{-x} - x^8e^{-x}$$
Setting this equal to zero, we get:
$$8x^7e^{-x} - x^8e^{-x} = 0$$
$$x^7e^{-x}(8-x)=0$$
This gives us two critical points: $x=0$ and $x=8$. 
To confirm that these are points of horizontal tangent lines, we can check the sign of $f''(x)$ at each critical point. Using the quotient rule:
$$f''(x) = 56x^6e^{-x} - 112x^5e^{-x} + x^8e^{-x}$$
At $x=0$, we have $f''(0) = 0$. This means that the tangent line at $x=0$ is either horizontal, vertical, or undefined. We can check this by looking at the original function: $f(0) = 0$, so the tangent line must be horizontal. 
At $x=8$, we have $f''(8) > 0$, which means that $x=8$ is a minimum point. Therefore, the tangent line at $x=8$ is also horizontal. 
So the answer is (C): (0,0), $(8, \frac{8^8}{e^8})$.

Find the Point(s) on the Graph of F Where the Tangent