Assignment 35

Good work.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

......!!!!!!!!...................................

21:50:42 5.2.18. Horiz line test, looks like log. What did the horizontal line test tell you for this function?

......!!!!!!!!...................................

RESPONSE --> the function is one-to-one.

.................................................

......!!!!!!!!...................................

21:50:52 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.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

21:51:37 5.2.20. Horiz line test, looks like inverted parabola or hyperbola. What did the horizontal line test tell you for this function?

......!!!!!!!!...................................

RESPONSE --> this is not a one-to-one function because the horizontal line intersects the graph at more than point.

.................................................

......!!!!!!!!...................................

21:51:46 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.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

21:53:49 5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse. Describe your sketch of the inverse function.

......!!!!!!!!...................................

RESPONSE --> the inverse function still has the points (1,1) and (-1,-1) but it intersects from the other side of the line y=x. It is a mirror of the graph of the function shown in the problem.

.................................................

......!!!!!!!!...................................

21:54:11 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.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

21:56:08 5.2.32 f = 2x + 6 inv to g = 1 / 2 * x – 3. Show that the functions f(x) and g(x) are indeed inverses.

......!!!!!!!!...................................

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

.................................................

......!!!!!!!!...................................

21:56:20 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.

......!!!!!!!!...................................

RESPONSE --> right

.................................................

......!!!!!!!!...................................

22:00:11 5.2.44. inv of x^3 + 1; domain range etc.. Give the inverse of the given function and the other requested information.

......!!!!!!!!...................................

RESPONSE --> f(x) = x^3 + 1 interchange x and y and solve for y x = y^3 + 1 y^3 = x -1 y = cbrt(x-1) f-1 = cbrt(x-1) f-1(f(x)) = f-1(x^3 + 1) = cbrt ((x^3 + 1) - 1) = x f(f-1(x)) = f(cbrt(x-1)) = (cbrt(x-1))^3 + 1 = x domain: (-inf, inf) range (-inf, inf)

.................................................

......!!!!!!!!...................................

22:00:39 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.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

22:12:07 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?

......!!!!!!!!...................................

RESPONSE --> the domain of f is (0, inf) since division by 0 isn't defined. the inverse of the function is: y=(3x + 1)/-x x = (3y+1)/-y x(-y)=3y + 1 x - 4y = 1 -4y = 1/x y = -1/4x

.................................................

......!!!!!!!!...................................

22:16:47 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.

......!!!!!!!!...................................

RESPONSE --> I see I did my algebra wrong here. How did you get (-x - 3)y when you factored out the y? I get -x * -y(-1 +3). I thought when you factored out you could only factor the common term (y).

-x * -y(-1 +3) = -x * -y (2) = 2 x y, which is not the same as -x y - 3 y.

On the other hand y ( -x - 3) = - x y - 3 y, by the distributive law.

Factoring in this example is just a matter of reversing the distributive law.

Let me know if this doesn't clarify the situation. I'll be glad to explain further if necessary.

.................................................

......!!!!!!!!...................................

22:18:46 5.2.74. T(L) = 2 pi sqrt ( L / g). Find L(T).

......!!!!!!!!...................................

RESPONSE --> g = 32.2 L(T) = (sqrt L/32.2)/(2pi)

.................................................

......!!!!!!!!...................................

22:21:32 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).

......!!!!!!!!...................................

RESPONSE --> Where did 6 in 6/(4pi^2) come from?

That's a typo. Should read g / (4 pi^2). I apparently typed the g as 6.

.................................................

"