Question

Let g be the function given by g(x) = x4 -4 x 3 +6x2 -4 x + k, where k is a constant.

A. On what intervals is g increasing? Justify your answer.
B. On what intervals is g concave upward? Justify your answer.
C. Find the value of k for which g has 5 as its relative minimum. Justify your answer.

Answer 1

A. To find the intervals on which g is increasing, we need to find the derivative of g and determine where it is positive. Taking the derivative of g, we get:

g'(x) = 4x^3 - 12x^2 + 12x - 4

To find the critical points, we set g'(x) = 0 and solve for x:

4x^3 - 12x^2 + 12x - 4 = 0

Dividing by 4, we get:

x^3 - 3x^2 + 3x - 1 = 0

(x - 1)^3 = 0

So x = 1 is the only critical point. To determine where g is increasing, we need to test a value in each of the intervals (-∞, 1) and (1, ∞). For example, if we plug in x = 0, we get:

g'(0) = -4 < 0

So g is decreasing on the interval (-∞, 1). If we plug in x = 2, we get:

g'(2) = 20 > 0

So g is increasing on the interval (1, ∞). Therefore, g is increasing on the interval (1, ∞).

B. To find the intervals on which g is concave upward, we need to find the second derivative of g and determine where it is positive. Taking the derivative of g', we get:

g''(x) = 12x^2 - 24x + 12

To find the critical points, we set g''(x) = 0 and solve for x:

12x^2 - 24x + 12 = 0

Dividing by 12, we get:

x^2 - 2x + 1 = 0

(x - 1)^2 = 0

So x = 1 is the only critical point. To determine where g is concave upward, we need to test a value in each of the intervals (-∞, 1) and (1, ∞). For example, if we plug in x = 0, we get:

g''(0) = 12 > 0

So g is concave upward on the interval (-∞, 1). If we plug in x = 2, we get:

g''(2) = 12 > 0

So g is concave upward on the interval (1, ∞). Therefore, g is concave upward on the intervals (-∞, 1) and (1, ∞).

C. To find the value of k for which g has 5 as its relative minimum, we need to set g'(x) = 0 and solve for x:

4x^3 - 12x^2 + 12x - 4 = 0

Dividing by 4, we get:

x^3 - 3x^2 + 3x - 1 = 0

(x - 1)^3 = 0

So x = 1 is the only critical point. To find the corresponding value of k, we plug in x = 1 and set g(1) = 5:

g(1) = 1^4 - 4(1)^3 + 6(1)^2 - 4(1) + k = 5

Simplifying, we get:

k = 6

Therefore, the value of k for which g has 5 as its relative minimum is 6.

Answer 2

A. Increasing intervals is (1, 2)

B. Concave upward interval is (1, 1.5)

C. Value of k for relative minimum at x = 1 5

A. Increasing Intervals

To determine the intervals where g is increasing, we need to analyze its derivative, g'(x).

g'(x) = 4x(x - 1)(x - 2)

g'(x) = 0 for x = 0, 1, 2

These critical points divide the number line into four intervals: (-∞, 0), (0, 1), (1, 2), and (2, ∞).

Evaluating g' at each interval:

(-∞, 0): g'(x) > 0, so g is increasing.

(0, 1): g'(x) < 0, so g is decreasing.

(1, 2): g'(x) > 0, so g is increasing.

(2, ∞): g'(x) < 0, so g is decreasing.

Therefore, g is increasing on the intervals (-∞, 0) and (1, 2).

B. Concave Upward Intervals

To determine the intervals where g is concave upward, we need to analyze its second derivative, g''(x).

g''(x) = 12(x - 1)(2x - 3)

g''(x) = 0 for x = 1, 3/2

These critical points divide the number line into three intervals: (-∞, 1), (1, 3/2), and (3/2, ∞).

Evaluating g'' at each interval:

(-∞, 1): g''(x) > 0, so g is concave upward.

(1, 3/2): g''(x) < 0, so g is concave downward.

(3/2, ∞): g''(x) > 0, so g is concave upward.

Therefore, g is concave upward on the intervals (-∞, 1) and (3/2, ∞).

C. Value of k for Relative Minimum

To find the value of k for which g has 5 as its relative minimum, we need to identify the x-value where g'(x) = 0 and g''(x) > 0. Since g'(x) = 0 for x = 0, 1, 2, and g''(x) > 0 for x = 1 and 3/2, only x = 1 satisfies both conditions.

To confirm that g has a relative minimum at x = 1, we can evaluate g' at x = 0, 1, and 2:

g'(0) > 0

g'(1) = 0

g'(2) > 0

Since g' changes from positive to zero and then back to positive, g has a relative minimum at x = 1.

To find the corresponding value of k, we substitute x = 1 into the original function:

g(1) = k

Therefore, k = 5 for g to have a relative minimum at x = 1.