Suppose f'(x) = xe^-x^2.

a) On what interval is f increasing? On what interval is f decreasing?

b) Does f have a maximum value? Minimum value?

So should I assume that f'(x) > 0 since e has to be a positive number?

Well, assuming f'(x) is never equal to zero, we can determine whether f is increasing or decreasing on different intervals based on the sign of f'(x). In this case, f'(x) = xe^(-x^2), so let's break it down.

a) To figure out where f is increasing or decreasing, we need to look at the sign of f'(x). Since e^(-x^2) is always positive, the only factor affecting the sign is x. When x is negative, f'(x) will also be negative. When x is positive, f'(x) will be positive. Therefore, f is decreasing on the interval (-∞, 0) and increasing on the interval (0, +∞).

b) Since f is always increasing on the entire range, it means there is no maximum value. As for a minimum value, there isn't one either since f continues to rise indefinitely. So, no max, no min. Just a fun ride on the upward slope!

Yes, you are correct. In order for f(x) to be increasing, f'(x) must be greater than zero. Since e^-x^2 is always positive, we can assume that if f'(x) = xe^-x^2 > 0, then x > 0. However, we need to determine the interval in which f(x) is increasing.

To do this, we need to find the critical points of f(x) by setting f'(x) = 0 and solving for x:

f'(x) = xe^(-x^2) = 0

Setting xe^-x^2 = 0, we find that x = 0 is the only critical point.

To determine the different intervals where f(x) is increasing or decreasing, we can choose test points in each interval and check the positivity/negativity of f'(x).

For the interval (-∞, 0), we can choose x = -1 as a test point. Evaluating f'(-1) = -e, we see that f'(-1) < 0. Therefore, f(x) is decreasing on the interval (-∞, 0).

For the interval (0, +∞), we can choose x = 1 as a test point. Evaluating f'(1) = e, we see that f'(1) > 0. Therefore, f(x) is increasing on the interval (0, +∞).

b) Since f(x) is increasing on the interval (0, +∞), it does not have a maximum value. However, since f(x) is decreasing on the interval (-∞, 0), it also does not have a minimum value.

To determine the intervals on which the function f(x) is increasing or decreasing, we can examine the sign of the derivative f'(x). However, in this case, assuming that f'(x) > 0 because e is positive is not accurate since x is also a variable.

Instead, to find the intervals of increasing or decreasing, we need to find the critical points of the function, where f'(x) = 0 or does not exist. Let's solve for the critical points first.

To find the critical points, set f'(x) equal to zero and solve for x:
xe^(-x^2) = 0.

Since e^(-x^2) is never equal to zero, the derivative f'(x) cannot be zero. Therefore, there are no critical points to consider.

Next, we can evaluate the sign of the derivative to determine the intervals of increasing or decreasing.

For f'(x) = xe^(-x^2):
- When x < 0, both x and e^(-x^2) are negative. Therefore, f'(x) < 0.
- When x > 0, both x and e^(-x^2) are positive. Thus, f'(x) > 0.

Based on these signs, we can conclude:

a) On the interval (-∞, 0), f(x) is decreasing.
On the interval (0, +∞), f(x) is increasing.

To determine if f(x) has a maximum or minimum value, we need to examine the behavior of f(x) as x approaches positive or negative infinity.

As x approaches negative infinity, e^(-x^2) approaches infinity, and f'(x) also approaches infinity. This indicates that f(x) does not have a maximum value.

As x approaches positive infinity, e^(-x^2) approaches zero, and f'(x) approaches zero as well. Therefore, f(x) does not have a minimum value either.

b) In conclusion, the function f(x) does not have a maximum or minimum value.

Remember, the assumptions about positive or negative values of variables need to be justified with solid reasoning based on the properties and relationships of the functions involved.

f'(x) > 0 means f is increasing

f' = (1-2x^2) e^-x^2
since e^z > 0 for all z, f is increasing where 1-2x^2 > 0, or -1/√2 < x < 1/√2

everywhere else it's decreasing

since f'(1/√2) = 0, and f changes from increasing to decreasing there, iot's a max.

similarly for min.