I have the solutions provide for the following questions, I would just like to a

Question

I have the solutions provide for the following questions, I would just like to ask for an expert to, step by step, explain to me how this solution came about and how does one reach this solution from the very start. Thanks!

2. Let f be function from U to U. (a) (5 points) Prove or disprove that f u f-1 is a function from U to U. Solution: Let f be the function {(a,c), (b, c)} then fuf-1 = {(a, b), (b, c), (c, a), (c, b)} and this is not a function. (b) (5 points) Prove or disprove that f;f-1 is a function from U to U. Solution: Let f be the function {a,c), (b, c)} then f;f 1 = {(a, a), (a, b), (6,5), (b, a)} and this is not a function. 3. Consider the real numbers R. Let f and g be the following two functions from R to R: f g = = {(x, y) y = x2 – 1} {(x, y)]y = 3} (a) (5 points) Prove or disprove that f-1 is a function. Solution: Observe that (0, 1) and (0, -1) are in f-1 and so f-1 is not a function. (b) (5 points) Determine f n g. Solution: f n g = {(2, 3), (–2, 3)}

Explanation / Answer

2.a) To solve this they have taken a function f as {(a,c) , (b,c)} means for x value of a y value is c and for x value of b y value is c.

Now the inverse of f will contain the pairs inversed i.e y value will become x value and x value will become y value so, f-1 will contain the mappings {(c,a) , (c,b)}

Now if we union both we get the mappings as = {(a,b),(b,c),(c,a),(c,b)}

Now we know that to become a function every x value should map to an unique y value OR in formal words every element in the domian of the function should map to an unique element in the co-domain/range.

But here c is mapped to both a and b so, this is not a function.

2.b) Now here also they have taken the same example of f as {(a,c),(b,c)}

Now f-1 is also {(c,a),(c,b)} like above.

Now f;f-1 is nothing bot f(f-1(x)).

Now in f we have (a,c) and in f-1 we have (c,a) therefore in f;f-1 we will have (a,a) as c is common.

In f we have (a,c) and in f-1 we have (c,b) therefore in f;f-1 we will have (a,b) as c is common.

In f we have (b,c) and in f-1 we have (c,a) therefore in f;f-1 we will have (b,a) as c is common.

In f we have (b,c) and in f-1 we have (c,b) therefore in f;f-1 we will have (b,b) as c is common.

Therefore in f;f-1 we have {(a,a),(a,b),(b,a),(b,b)}

Now as a is mapped to two values a and b and b is also mapped to two values a and b so f;f-1 is not a function.

3 a) Now f is given as y= x2 - 1

or, x2 = y + 1

or, x = +-sqrt(y + 1)

Therefore f-1(x) = +-sqrt(x + 1)

Now if we put x=0

we get two values of  f-1(x) 1 and -1.

Therefore 0 maps to two values 1 and -1 so, f-1 is not a function.

3 b) g is defined by y=3

and f is y=x2 - 1.

Putting y=3 in f we got:-

3=x2 - 1

x2 = 4

x=+2 and x=-2

Therefore intersection of f and g will have {(2,3),(-2,3)}

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