Alicia is solving the equation shown.
1 answer:
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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
We want to find the mean of two elements in a set, given that we know the other elements of the set and the mean of the whole set.
The answer is: -490
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For a set with N elements {x₁, x₂, ..., xₙ} the mean is given by:
Here we know that:
- The mean of the set is 0.
- The set has 1000 elements.
- 998 of these elements are ones, the other two are A and B.
We want to find the mean of the values of A and B.
First, we can start by writing the equation for the mean:
We can rewrite this as:
And we have 998 ones, then:
Now we have B isolated.
With this, the mean of A and B can be written as:
So we can conclude that the mean of the other two numbers is -490.
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