Analytical Applications of Differentiation unit test
A test on this unit alone, marked as a percentage and a letter grade — for the test your class is actually sitting, rather than for May. Answer everything, then submit once: seeing the answer to question 3 before attempting question 4 makes the final percentage meaningless.
Even power blocks a sign change
Sign information gives shape, not height
Second derivative test is inconclusive at zero
Two-row sign chart
Justification names a derivative
Optimization procedure
Absolute extremum justification
Reading f′ to describe f
Inflection point needs a sign change
Point of inflection
Justification language
Because the graph turns around
Short answer 1. Define or explain: Candidates Test procedure
3 ptsShort answer 2. Define or explain: Forgetting the endpoints
3 ptsShort answer 3. Define or explain: First derivative test
3 ptsShort answer 4. Define or explain: Mean Value Theorem hypotheses
3 ptsFree response
9 ptsNO CALCULATOR. The continuous function f is defined on the closed interval −6 ≤ x ≤ 12. The graph of f consists of two semicircles and one line segment: • a semicircle below the x-axis from (−6, 0) to (0, 0), centered at (−3, 0) with radius 3 (minimum value −3 at x = −3); • a semicircle above the x-axis from (0, 0) to (6, 0), centered at (3, 0) with radius 3 (maximum value 3 at x = 3); • a line segment from (6, 0) to (12, 3). Let g be the function defined by g(x) = ∫₆ˣ f(t) dt.
A. Find g′(8). Give a reason for your answer.
B. Find all values of x in the open interval −6 < x < 12 at which the graph of g has a point of inflection. Give a reason for your answer.
C. Find g(12) and g(0). Label your answers.
D. Find the value of x at which g attains an absolute minimum on the closed interval −6 ≤ x ≤ 12. Justify your answer.