Question

NO CALCULATOR IS ALLOWED FOR THIS QUESTION.

Show all of your work, even though the question may not explicitly remind you to do so. Clearly
label any functions, graphs, tables, or other objects that you use. Justifications require that you give
mathematical reasons, and that you verify the needed conditions under which relevant theorems,
properties, definitions, or tests are applied. Your work will be scored on the correctness and
completeness of your methods as well as your answers. Answers without supporting work will
usually not receive credit.
Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is
given as a decimal approximation, it should be correct to three places after the decimal point.
Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers
x for which f(x) is a real number.
f(x) =
Let f be the function defined above.
√9-x²
for -3≤x≤0
-x+3 cos (pie*x/2) for 0 < x≤ 4
(a) Find the average rate of change of f on the interval -3 ≤x ≤ 4.

(b) Write an equation for the line tangent to the graph of f at x= 3.
(c) Find the average value of f on the interval-3 ≤x≤ 4.

(d) Must there be a value of x at which f(x) attains an absolute maximum on the closed interval -3 ≤x≤ 4 Justify your answer.

Answer 1

(a) The average rate of change of f on the interval [-3,4] is given by:

(1/(4-(-3))) * ∫[a,b] f(x) dx

where a = -3 and b = 4. We can break up the integral into two parts, one over the interval [-3,0] and the other over the interval (0,4]:

(1/7) * [∫[-3,0] √(9-x²) dx + ∫[0,4] (-x+3cos(πx/2)) dx]

For the first integral, we recognize that the integrand is the equation of the top half of a circle with radius 3 centered at the origin. Therefore, we can use the substitution x = 3sin(t), dx = 3cos(t)dt, to get:

∫[-3,0] √(9-x²) dx = ∫[-π/2,0] 9cos²(t) dt = (9/2) * [sin(t)cos(t) + t]_[-π/2,0] = (9π - 81)/4

For the second integral, we can use integration by parts with u = -x and dv = cos(πx/2) dx to get:

∫[0,4] (-x+3cos(πx/2)) dx = [-x²/2 + (6/π)sin(πx/2)]_0^4 = -8

Therefore, the average rate of change of f on the interval [-3,4] is:

(1/7) * [(9π - 81)/4 - 8] = (9π - 145)/28

(b) To find the equation of the tangent line to the graph of f at x = 3, we need to find the slope of the tangent. Since f is not differentiable at x = 0 (due to the cosine term), we need to consider the left and right derivatives separately.

For x < 0, the function is the equation of the top half of a circle with radius 3 centered at the origin, so the slope of the tangent at x = 3 is:

f'(3-) = -√(9-3²)/(3-0) = -√6

For x > 0, we have:

f'(x) = -1 - (3π/4)sin(πx/2)

So the slope of the tangent at x = 3 is:

f'(3+) = -1 - (3π/4)sin(3π/2) = -1 + (3π/4)

The equation of the tangent line is therefore:

y - f(3) = f'(3)(x-3)

y + √(9-3²) = (-√6)(x-3) (for x < 0)

y - 6 + 3cos(π/2) = [(-1 + (3π/4))(x-3)] (for x > 0)

(c) The average value of f on the interval [-3,4] is given by:

(1/(4-(-3))) * ∫[-3,4] f(x) dx

Using the same breakdown of the integral as in part (a), we have:

(1/7) * [∫[-3,0] √(9-x²) dx + ∫[0,4] (-x+3cos(πx/2)) dx]

The first integral was evaluated in part (a

Explanation: