Final answer:
The critical values of f(x) are x = 0 and x = e^(-1/2). f(x) is increasing on the intervals (-∞, 0) and (e^(-1/2), ∞) and decreasing on the interval (0, e^(-1/2)). The local maxima of f(x) occur at x = 0, and the local minima occur at x = e^(-1/2). f(x) is concave up on the interval (0, ∞) and concave down on the interval (-∞, 0). The inflection point of f(x) occurs at x = 0.
Step-by-step explanation:
(a) To find the critical values of f(x), we need to find the values of x where the derivative of f(x) is equal to zero or undefined. The derivative of f(x) is given by f'(x) = 12xln(x) + 6x. Setting this equal to zero, we can solve for x:
12xln(x) + 6x = 0
6x(2ln(x) + 1) = 0
So, x = 0 or 2ln(x) + 1 = 0 => ln(x) = -1/2 => x = e^(-1/2). Therefore, the critical values of f(x) are x = 0 and x = e^(-1/2).
(b) To determine where f(x) is increasing, we need to find the intervals where the derivative f'(x) is positive. Since f'(x) = 12xln(x) + 6x, we look for the values of x that make f'(x) positive. From our analysis in part (a), we know that x = 0 is a critical value, so we can consider x < 0 and x > 0 separately. If x < 0, then 12xln(x) + 6x is negative. If x > e^(-1/2), then 12xln(x) + 6x is positive. Therefore, f(x) is increasing on the intervals (-∞, 0) and (e^(-1/2), ∞).
(c) To determine where f(x) is decreasing, we need to find the intervals where the derivative f'(x) is negative. From our analysis in part (b), we know that f(x) is increasing on the intervals (-∞, 0) and (e^(-1/2), ∞). Therefore, f(x) is decreasing on the interval (0, e^(-1/2)).
(d) To find the x values of all the local maxima of f(x), we need to find the values of x for which the derivative f'(x) changes from positive to negative. From our analysis in part (a), we know that x = 0 and x = e^(-1/2) are critical values. Since f(x) is increasing on the interval (-∞, 0) and decreasing on the interval (0, e^(-1/2)), we can conclude that x = 0 is a local maximum.
(e) To find the x values of all the local minima of f(x), we need to find the values of x for which the derivative f'(x) changes from negative to positive. From our analysis in part (a), we know that x = 0 and x = e^(-1/2) are critical values. Since f(x) is decreasing on the interval (0, e^(-1/2)) and increasing on the interval (e^(-1/2), ∞), we can conclude that x = e^(-1/2) is a local minimum.
(f) To find where f(x) is concave up, we need to find the intervals where the second derivative f''(x) is positive. The second derivative of f(x) is given by f''(x) = 12ln(x) + 12. Since ln(x) is only defined for x > 0, we can consider x > 0 separately. For x > 0, 12ln(x) + 12 is positive. Therefore, f(x) is concave up on the interval (0, ∞).
(g) To find where f(x) is concave down, we need to find the intervals where the second derivative f''(x) is negative. From our analysis in part (f), we know that f(x) is concave up on the interval (0, ∞). Therefore, f(x) is concave down on the interval (-∞, 0).
(h) To find the x values of all the inflection points of f, we need to find the values of x where the second derivative f''(x) changes from positive to negative or from negative to positive. From our analysis in parts (f) and (g), we know that x = 0 is a critical value. Since f(x) is concave up on the interval (0, ∞) and concave down on the interval (-∞, 0), we can conclude that x = 0 is an inflection point.
(i) To sketch a graph of f, we need to consider the different intervals identified in parts (b), (c), (f), and (g), as well as the critical points and inflection point identified in parts (a) and (h). We can also determine the behavior of f(x) as x approaches ±∞. The vertical asymptote occurs at x = 0, as ln(x) approaches -∞ as x approaches 0 from the right. The horizontal asymptote occurs at y = ±∞, as f(x) approaches ±∞ as x approaches ±∞.