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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
Answer:
The coordinates of the circumcenter of this triangle are (3,2)
Step-by-step explanation:
we know that
The circumcenter is the point where the perpendicular bisectors of a triangle intersect
we have the coordinates
step 1
Find the midpoint AB
The formula to calculate the midpoint between two points is equal to
substitute the values
step 2
Find the equation of the line perpendicular to the segment AB that passes through the point (-2,2)
Is a horizontal line (parallel to the x-axis)
-----> equation A
step 3
Find the midpoint BC
The formula to calculate the midpoint between two points is equal to
substitute the values
step 4
Find the equation of the line perpendicular to the segment BC that passes through the point (3,-1)
Is a vertical line (parallel to the y-axis)
-----> equation B
step 5
Find the circumcenter
The circumcenter is the intersection point between the equation A and equation B
-----> equation A
-----> equation B
The intersection point is (3,2)
therefore
The coordinates of the circumcenter of this triangle are (3,2)