Answer:
Jackson should work at least 8 hours this week to achieve his target.
Step-by-step explanation:
Jackson's hourly income = $25
His weekly spending budget = $130
Minimum amount to be pushed to the savings account = $50
So, minimum total amount Jackson should earn this week = 130 + 50 = 180
Number of hours to work this week =
=
= 8 rounded to next integer
Hence, Jackson should work at least 8 hours this week to achieve his target.
Answer:
a. 5 b. c. 148.5 d. 1/7
Step-by-step explanation:
Here is the complete 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 for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.
Solution
a. f"(2)
f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x
f"(2) = cos(π × 2) + 2 × 2
f"(2) = cos(2π) + 4
f"(2) = 1 + 4
f"(2) = 5
b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0
We first find f(x) by integrating f'(x)
f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C
f(0) = 2 so,
2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C
2 = -cos(0)/π + 0 + 0 + C
2 = -1/π + C
C = 2 + 1/π
f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π
f(x) = [1-cos(πx)]/π + x³/3 +3x + 2
y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)
The tangent to y is thus dy/dx
dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx
dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)
at x = 0,
dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)
dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)
dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)
dy/dx = -(0/π + 2)⁻²(3)
dy/dx = -(0 + 2)⁻²(3)
dy/dx = -(2)⁻²(3)
dy/dx = -3/4
At x = 0,
y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)
y = 1/([1-cos(0)]/π + 0 + 0 + 2)
y = 1/([1 - 1]/π + 2)
y = 1/(0/π + 2)
y = 1/(0 + 2)
y = 1/2
So, the equation of the tangent at (0, 1/2) is
c. If g(x) = f (√(3x² + 4). Find g'(2)
g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2
g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)
g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)
g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)
g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)
g'(2) = [6sin4π + 576 + 18]/4)
g'(2) = [6 × 0 + 576 + 18]/4)
g'(2) = [0 + 576 + 18]/4)
g'(2) = 594/4
g'(2) = 148.5
d. If h be the inverse function of f. Find h' (2)
If h(x) = f⁻¹(x)
then h'(x) = 1/f'(x)
h'(x) = 1/(sin(πx) + x² +3)
h'(2) = 1/(sin(π2) + 2² +3)
h'(2) = 1/(sin(2π) + 4 +3)
h'(2) = 1/(0 + 4 +3)
h'(2) = 1/7