Question

If f and g are the functions whose graphs are shown, let u(x) = f(g(x)), v(x) = g(f(x)), and w(x) = g(g(x)). Find each derivative, if it exists. If it does not exist, explain why. (If an answer does not exist, enter DNE.) The x y-coordinate plane is given. There are two functions on the graph. The function labeled f enters the window in the second quadrant, goes down and right, sharply changes direction at the origin, goes up and right, passes through the point (1, 3), sharply changes direction at the point (3, 9), goes down and right, passes through the point (7, 8), and exits the window in the first quadrant. The function labeled g enters the window at approximately x = −2.3 on the negative x-axis, goes up and right, sharply changes direction at the point (0, 7), goes down and right, passes through the point (1, 6), sharply changes direction at the point (4, 3), goes up and right, passes through the point (7, 4), and exits the window in the first quadrant. (a) u ′(1) = 0 It does exist. u'(1) does not exist because f ′(6) does not exist. u'(1) does not exist because g′(1) does not exist. u'(1) does not exist because f ′(7) does not exist. u'(1) does not exist because g′(6) does not exist. Correct: Your answer is correct. (b) v ′(1) = 2 Incorrect: Your answer is incorrect. correct It does exist. v'(1) does not exist because f ′(1) does not exist. v'(1) does not exist because g′(1) does not exist. v'(1) does not exist because f ′(3) does not exist. v'(1) does not exist because g′(3) does not exist. Incorrect: Your answer is incorrect. (c) w ′(1) = -2 Incorrect: Your answer is incorrect. It does exist. w'(1) does not exist because f ′(1) does not exist. w'(1) does not exist because g′(1) does not exist. w'(1) does not exist because f ′(6) does not exist. w'(1) does not exist because g′(6) does not exist.

Answer 1

(a) u'(1) does not exist because g′(1) does not exist.

(b) v'(1) does not exist because g′(1) does not exist.

(c) w'(1) does not exist because f ′(1) does not exist.

Step-by-step explanation:

(a) To find u'(1), we need to use the chain rule, which states that if u(x) = f(g(x)), then u'(x) = f'(g(x)) * g'(x). Evaluating at x = 1, we find that g′(1) does not exist, leading to u'(1) not existing.

(b) For v'(1), applying the chain rule again, we get v'(x) = g'(f(x)) * f'(x). Evaluating at x = 1, we find that g′(1) does not exist, resulting in v'(1) not existing.

(c) To find w'(1), using the chain rule, we have w'(x) = g'(g(x)) * g'(x). Evaluating at x = 1, we find that f ′(1) does not exist, leading to w'(1) not existing.

In each case, the non-existence of the derivative is determined by the fact that at x = 1, either f ′(1) or g′(1) or both do not exist. The chain rule relies on the existence of the derivatives of the functions involved, and if any of those derivatives is non-existent at the specified point, then the derivative of the composite function will also not exist at that point. Therefore, the correct answers are provided based on the non-existence of the relevant derivatives.