Assignment 35

course Mth 158

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12:22:39

5.2.18. Horiz line test, looks like log.

What did the horizontal line test tell you for this function?

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Horizontal line test passes through the graph once, therefore the function is one to one.

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12:22:47

There is no horizontal line that passes through this graph more than once. The function is strictly increasing, taking each y value only once. The function is therefore one-to-one on its domain.

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12:23:18

5.2.20. Horiz line test, looks like inverted parabola or hyperbola.

What did the horizontal line test tell you for this function?

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Horizontal line test passes through the graph twice, therefore the function is not one to one.

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12:23:21

For every horizontal below the 'peak' of this graph the graph will intersect the horizontal line in two points. This function is not one-to-one on the domain depicted here.

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12:27:09

5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse.

Describe your sketch of the inverse function.

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Moving left to right, in the third quadrant the function is increasing and an increasing rate, passes through the origin and into the first quadrant increasing at a decreasing rate.

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12:27:12

The graph of the function passes through (0, 0), (1,1), and (-1,-1). The inverse function will reverse these coordinates, which will give the same three points.

Between x = -1 and x = 1 the graph of the original function is closer to the x axis than to the y axis, and is horizontal at the origin. The graph of the inverse function will therefore be closer to the y axis than to the x axis for y values between -1 and 1, and will be vertical at the origin.

For x < 1 and for x > 1 the graph lies closer to the y axis than to the x axis. The graph of the inverse function will therefore lie closer to the x axis than to the y axis for y < 1 and for y > 1.

In the first quadrant the function is increasing at an increasing rate. The inverse function will therefore be increasing at a decreasing rate in the first quadrant.

In the third quadrant the function is increasing at a decreasing rate. The inverse function will therefore be increasing at an increasing rate in the third quadrant.

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12:29:27

f(g(x)) = 2 g(x) + 6 = 2 ( 1 / 2 * x - 3) + 6 = x - 6 + 6 = x.

g(f(x)) = 1 / 2 * f(x) - 3 = 1/2 ( 2 x + 6) - 3 = x + 3 - 3 = x.

Since f(g(x)) = g(f(x)) = x, the two functions are inverse.

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f(x) = 2x + 6 and g(x) = 1/2x - 3

f(g(x)) = 2(1/2x - 3) + 6 = x - 6 + 6 = x

g(f(x) = 1/2 (2x + 6) - 3 = x + 3 - 3 = x

Functions are inverse.

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12:31:15

5.2.44. inv of x^3 + 1; domain range etc..

Give the inverse of the given function and the other requested information.

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f(x) = x^3 + 1

y = x^3 + 1

x = y^3 + 1

y^3 = x - 1

y = x - 1 ^(1/3)

f-1(x) = x - 1 ^(1/3)

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12:31:24

The function is y = x^3 + 1. This function is defined for all real-number values of x and its range consists of all real numbers.

If we switch the roles of x and y we get x = y^3 + 1. Solving for y we get

y = (x - 1)^(1/3).

This is the inverse function. We can take the 1/3 power of any real number, positive or negative, so the domain of the inverse function is all real numbers. Any real-number value of y can be obtained by using an appropriate value of x. So both the domain and range of the inverse function consist of all real numbers.

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12:35:23

5.2.56. inv of f(x) = (3x+1)/(-x). Domain and using inv fn range of f.

What is the domain of f? What is the inverse function? What does the inverse function tell you about the range of f?

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f(x) = (3x+1)/(-x)

x = (3y+1)/(-y)

-xy = 3y + 1

-xy - 3y = 1

y(-x - 3) = 1

y(-x - 3)/-x - 3 = 1/-x - 3

y = 1/(-x - 3)

f-1(x) = -1/(x + 3)

Domain of f = reange of f-1(x) = {x|x not equal 0}

Range of f = Domain of f-1(x) = {x|x not equal 3}

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12:35:32

f(x) is defined for all x except x = 0, since division by 0 is not defined.

If we switch x and y in the expression y = (3x + 1) / (-x) we get x = (3y + 1) / (-y). To solve for y we first multiply by -y, noting that this excludes y = 0 since multiplication of both sides by 0 would change the solution set. We get

-x y = 3y + 1. Subtracting 3 y from both sides we get

-x y - 3 y = 1. Factoring y out of the left-hand side we get

(-x - 3) y = 1, and dividing both sides by (-x - 3), which excludes x = -3, we get

y = -1 / (x + 3).

The domain of this function is the set of all real numbers except 3. Since the domain of the inverse function is the range of the original function, the range of the original function consists of all real numbers except 3.

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12:38:28

5.2.74. T(L) = 2 pi sqrt ( L / g). Find L(T).

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T(L) = 2 pi sqrt ( L / g)

(t)^2 = (2 pi sqrt ( L / g))^2

t^2 = 4pi^2 * L/g

g/4pi^2(T^2) = 4pi^2 * L/g(g/4pi^2)

L = T^2*g/4pi^2

L(T) = T^2*g/4pi^2

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12:38:32

We solve T = 2 pi sqrt( L / g) for L. First squaring both sides we obtain

T^2 = 4 pi^2 * L / g. Multiplying both sides by 6 / ( 4 pi^2) we get

L = T^2 g / (4 pi^2).

So our function L(T) is

L(T) = T^2 g / (4 pi^2).

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Assignment 35

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