Concavity and Function Behavior Quiz

Determine the concavity of the function f(x) = 3x^2 - 6x + 2.

• The function f(x) = 3x^2 - 6x + 2 has no concavity.

• The function f(x) = 3x^2 - 6x + 2 is concave down.

• The function f(x) = 3x^2 - 6x + 2 is linear.

• The function f(x) = 3x^2 - 6x + 2 is concave up.

Find the concavity of the function g(x) = x^3 - 6x^2 + 9x - 4.

• The concavity of the function g(x) = x^3 - 6x^2 + 9x - 4 is downward (or concave down).

• The concavity of the function g(x) = x^3 - 6x^2 + 9x - 4 is linear.

• The concavity of the function g(x) = x^3 - 6x^2 + 9x - 4 is upward (or concave up).

• The concavity of the function g(x) = x^3 - 6x^2 + 9x - 4 is undefined.

Tags

CCSS.HSF.IF.A.2

Analyse the behavior of the function h(x) = 4x^3 - 12x^2 + 8x.

• The function h(x) = 4x^3 - 12x^2 + 8x has no specific behavior

• The function h(x) = 4x^3 - 12x^2 + 8x is always decreasing

• The function h(x) = 4x^3 - 12x^2 + 8x is always increasing

• The function h(x) = 4x^3 - 12x^2 + 8x exhibits behavior of increasing from negative infinity to 1, then decreasing from 1 to 2, and increasing from 2 to positive infinity.

Examine the behavior of the function k(x) = -2x^3 + 6x^2 - 4x.

• The function k(x) = -2x^3 + 6x^2 - 4x is a linear function, so it will have a constant end behavior for all x values.

• The function k(x) = -2x^3 + 6x^2 - 4x is an exponential function, so it will have end behavior that oscillates as x approaches infinity.

• The function k(x) = -2x^3 + 6x^2 - 4x is a quadratic function, so it will have end behavior that approaches negative infinity as x approaches positive infinity.

• The function k(x) = -2x^3 + 6x^2 - 4x is a cubic function, so it will have end behavior that approaches negative infinity as x approaches negative infinity, and end behavior that approaches positive infinity as x approaches positive infinity.

Identify the intervals of increasing and decreasing for the function f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1.

• The intervals of increasing are (-∞, 1) and the intervals of decreasing are (1, ∞)

• The intervals of increasing are (-∞, 0) and the intervals of decreasing are (0, ∞)

• The intervals of increasing are (1, ∞) and the intervals of decreasing are (-∞, 1)

• The intervals of increasing are (-∞, 2) and the intervals of decreasing are (2, ∞)

Determine the intervals of increasing and decreasing for the function g(x) = -2x^4 + 8x^3 - 12x^2 + 8x.

• The function g(x) is increasing on the intervals (-∞, 2) and (3, ∞), and decreasing on the interval (2, 3)

• The function g(x) is increasing on the intervals (-∞, 2) and (3, ∞), and decreasing on the interval (2, 3)

• The function g(x) is increasing on the intervals (-∞, 0) and (1, ∞), and decreasing on the interval (0, 1)

• The function g(x) is increasing on the intervals (-∞, 1) and (2, ∞), and decreasing on the interval (1, 2).

Apply the first derivative test to find the local extrema of the function f(x) = 2x^3 - 6x^2 + 4x - 1.

• f(x) has local maxima at x = 2 and local minima at x = 1

• f(x) has local maxima at x = 1 and local minima at x = 2

• f(x) has local maxima at x = -1 and local minima at x = 2

• f(x) has local maxima at x = 0 and local minima at x = 3

Use the first derivative test to determine the local extrema of the function g(x) = -x^3 + 3x^2 - 2x + 5.

• The local extrema are at x = 1 and x = 2.

• The local extrema are at x = -1 and x = -2.

• The local extrema are at x = 4 and x = 5.

• The local extrema are at x = 0 and x = 3.

Apply the second derivative test to classify the critical points of the function f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1.

• The critical points are all points of discontinuity.

• The critical points are all inflection points.

• The critical points are classified as follows: f''(x) = 12x^2 - 24x + 12, f''(x) > 0 for all x, so the critical points are all local minima.

• The critical points are all local maxima.

Use the second derivative test to classify the critical points of the function g(x) = -2x^4 + 8x^3 - 12x^2 + 8x.

• The second derivative is g''(x) = -24x^2 + 48x - 24. The critical points are at x = 0 and x = 3. Evaluating g''(0) gives 24, which is greater than 0, indicating a local minimum.

• The second derivative is g''(x) = -24x^2 + 48x - 24. The critical points are at x = 1 and x = 2. Evaluating g''(1) gives -24, which is less than 0, indicating a local maximum. Evaluating g''(2) gives 24, which is greater than 0, indicating a local minimum.

• The critical points are at x = -1 and x = -2. Evaluating g''(-1) gives 24, which is greater than 0, indicating a local minimum.

• The second derivative is g''(x) = -24x^2 + 48x - 24. The critical points are at x = 3 and x = 4. Evaluating g''(3) gives -24, which is less than 0, indicating a local maximum.