Find the interval on which f(x)=xe^-x is increasing.
I got (1,infinity)..
and you would be wrong. You have it exactly backwards.
f is increasing where f' > 0.
Better double-check your derivative.
To determine the interval on which a function is increasing, we need to find the values of x for which the derivative of the function is positive.
Let's find the derivative of the function f(x) = x * e^(-x) using the product rule.
f'(x) = (1 * e^(-x)) + (x * (-1 * e^(-x)))
= e^(-x) - x * e^(-x)
= e^(-x)(1 - x)
To find the values of x for which f'(x) > 0, we need to solve the inequality e^(-x)(1 - x) > 0.
First, let's consider the sign of e^(-x). The exponential function e^(-x) is always positive since it represents the inverse of the natural exponential function, which is always positive.
Next, we consider the sign of (1 - x). For this part, let's set (1 - x) = 0 and solve for x:
1 - x = 0
x = 1
From this, we can see that (1 - x) is negative for x > 1 and positive for x < 1.
To determine the overall sign of e^(-x)(1 - x), we can create a sign chart:
x < 1 | x = 1 | x > 1
--------------------------------
(+) | 0 | (-)
From the sign chart, we can see that e^(-x)(1 - x) is positive for x < 1.
Therefore, the interval on which f(x) = x * e^(-x) is increasing is (-∞, 1).
So, the correct interval is (-∞, 1).