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IRINA_888 [86]
2 years ago
14

Algebra help pls ...i uploaded a picture

Mathematics
1 answer:
____ [38]2 years ago
7 0

Answer:

I'll give you some clues in the solution. Please read them so you can do the rest faster.

Solution:

First thing you do is write the vertex. The vertex is basically the last number in the absolute value bars and the number outside of them (to the right of the bars). The vertex is the opposite of the last number in the absolute value brackets and if there is no number to the right, assume it is 0.

We start at (2,0).

The domain is 99% of the time all real numbers. Write that for domain.

The range is the y value where things start to increase (the y value of the vertex). The range is 0.

The type of function is absolute value function btw

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Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tab
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Answer:

a. 5 b. y = -\frac{3}{4}x + \frac{1}{2} 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

\frac{y - \frac{1}{2} }{x - 0} = -\frac{3}{4}  \\y - \frac{1}{2} = -\frac{3}{4}x\\y = -\frac{3}{4}x + \frac{1}{2}

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

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How can I find the area of a rectangle with a rectangle the size of w- 2in and l- 5 in. I need two ways and i need a lot of easy
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Here's one way: A=WxL So in this case, it's 2x5 You would draw the rectangle, label one side 2 in. and the other 5 in. Draw an arrow from each no./side and put x underneath the line so that its 2 —x— 5 (2, line, x, line, 5. With x under line)
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