Consider the function f(x)=xlnx.

For the following questions, write inf for ∞, -inf for −∞, and None if no such answer exists.

a.) The domain of this function is .

b.) The x-intercept is .

c.) f′(x) = .

d.) f(x) is increasing on the interval and decreasing on .

e.) f(x) has a local minimum at and a local maximum at .

f.) f′′(x) = .

g.) f(x) is concave up on the interval and concave down on .

h.) f(x) has a point of inflection at .

To find the answers for these questions, let's go through each one step by step:

a) The domain of the function f(x) = xln(x) is the set of all real numbers greater than 0 because the natural logarithm ln(x) is only defined for positive values of x. Therefore, the domain is (0, ∞).

b) The x-intercept represents the value of x for which the function f(x) equals zero. In this case, we need to solve the equation xln(x) = 0. There are two solutions: x = 0 and x = 1. However, since ln(x) is not defined for x = 0, the x-intercept is x = 1.

c) To find the derivative f'(x), we can use the product rule. Let's differentiate the function term by term. The derivative of xln(x) with respect to x is:

f'(x) = x * d(ln(x))/dx + ln(x) * d(x)/dx.

Using the chain rule, we find that d(ln(x))/dx = 1/x and d(x)/dx = 1.

Therefore, f'(x) = x * (1/x) + ln(x) * 1 = 1 + ln(x).

d) To determine whether f(x) is increasing or decreasing on an interval, we need to analyze the sign of its derivative f'(x) on that interval. Let's consider the intervals (0, ∞) and (-∞, 0).

On the interval (0, ∞): Since f'(x) = 1 + ln(x), ln(x) is positive for x > 1, and therefore, f'(x) is positive for x > 1. We conclude that f(x) is increasing on (1, ∞).

On the interval (-∞, 0): Since f'(x) is not defined for values of x less than or equal to 0, we cannot determine whether f(x) is increasing or decreasing on this interval.

e) To find the local minimum and maximum, we need to analyze critical points. A critical point occurs when f'(x) = 0 or when f'(x) is not defined.

To find the critical point for f'(x) = 1 + ln(x), set 1 + ln(x) = 0 and solve for x:

ln(x) = -1,

x = e^(-1) ≈ 0.3679.

Now, we need to check the second derivative f''(x) to determine the nature of the critical point.

f''(x) represents the second derivative of f(x) and measures the concavity of the function.

f''(x) = d^2(ln(x))/dx^2 = 1/x.

f''(x) is positive for x > 0, which implies that f(x) is concave up on the interval (0, ∞). Since f''

(x) changes sign at x = e^(-1), we conclude that f(x) has a local minimum at x = e^(-1) and no local maximum.

f) The second derivative f''(x) = 1/x.

g) Similar to the previous question, we need to analyze the sign of the second derivative f''(x) on different intervals to determine the concavity of the function f(x).

On the interval (0, ∞): Since f''(x) = 1/x, f''(x) is positive for x > 0, indicating that the function f(x) is concave up on (0, ∞).

On the interval (-∞, 0): Since f''(x) is not defined for values of x less than or equal to 0, we cannot determine the concavity of the function on this interval.

h) A point of inflection occurs when the concavity of the function changes. In this case, since f(x) = xln(x), the function does not change concavity. Therefore, there are no points of inflection for this function.

To summarize:

a) The domain of f(x) = xln(x) is (0, ∞).
b) The x-intercept is x = 1.
c) f'(x) = 1 + ln(x).
d) f(x) is increasing on the interval (1, ∞), but we cannot determine whether it is increasing or decreasing on (-∞, 0).
e) f(x) has a local minimum at x = e^(-1) and no local maximum.
f) f''(x) = 1/x.
g) f(x) is concave up on the interval (0, ∞), but we cannot determine the concavity on (-∞, 0).
h) f(x) does not have any points of inflection.