What are the absolute extrema of #f(x)=(x^4) / (e^x) in[0,oo]#?

Question

Isabella Ackerman · Accepted Answer

The minimum is #0# at #x=0#, and the maximum is #4^4/e^4# at #x=4# Note first that, on #[0,oo)#, #f# is never negative. Furthermore, #f(0)=0# so that must be the minimum. #f'(x) = (x^3(4-x))/e^x# which is positive on #(0,4)# and negative on #(4,oo)#. We conclude that #f(4)# is a relative maximum. Since the function has no other critical points in the domain, this relative maximum is also the absolute maximum.

Ezekiel Alexander · Answer

undefined To find the absolute extrema of $ f(x) = \frac{x^4}{e^x} $ in the interval $[0, \infty)$, we need to analyze the critical points and the behavior of the function at the endpoints.

First, find the critical points by taking the derivative of $ f(x) $ and setting it equal to zero:
$ f'(x) = \frac{d}{dx} \left( \frac{x^4}{e^x} \right) = \frac{4x^3e^x - x^4e^x}{(e^x)^2} = \frac{x^3(4 - x)e^x}{e^{2x}} $.

Setting $ f'(x) = 0 $, we find critical points at $ x = 0 $ and $ x = 4 $.

Next, evaluate the function at the critical points and at the endpoints of the interval:
- $ f(0) = \frac{0^4}{e^0} = 0 $
- $ f(4) = \frac{4^4}{e^4} $

Since $ e^4 $ is a positive constant, maximizing $ \frac{x^4}{e^x} $ is equivalent to maximizing $ x^4 $. Therefore, the absolute maximum occurs at $ x = 4 $.

Since $ \lim_{x \to \infty} \frac{x^4}{e^x} = 0 $, there is no absolute minimum in the interval $[0, \infty)$.

Therefore, the absolute maximum of $ f(x) $ in the interval $[0, \infty)$ is $ \frac{4^4}{e^4} $.

Elijah Abram · Answer

undefined To find the absolute extrema of the function $ f(x) = \frac{x^4}{e^x} $ on the interval $[0, \infty)$, we first need to determine the critical points of the function within this interval. To do so, we find where the derivative of the function is equal to zero or undefined. Then, we evaluate the function at these critical points as well as at the endpoints of the interval to determine the absolute extrema.

Taking the derivative of $ f(x) $ with respect to $ x $, we get:

$$ f'(x) = \frac{d}{dx} \left( \frac{x^4}{e^x} \right) = \frac{4x^3e^x - x^4e^x}{(e^x)^2} = \frac{x^3(4 - x)e^x}{e^{2x}} $$

Setting $ f'(x) = 0 $, we find the critical points by solving the equation $ x^3(4 - x)e^x = 0 $. The solutions are $ x = 0 $ and $ x = 4 $.

Next, we evaluate the function $ f(x) $ at these critical points and at the endpoint $ x = 0 $ of the interval:

$$ f(0) = \frac{0^4}{e^0} = 0 $$
$$ f(4) = \frac{4^4}{e^4} \approx 15.439 $$

Since $ e^x $ increases as $ x $ increases, as $ x $ approaches infinity, $ f(x) $ approaches zero. Therefore, the absolute minimum occurs at $ x = 0 $ with a value of $ 0 $, and the absolute maximum occurs at $ x = 4 $ with a value of approximately $ 15.439 $.