34. Suppose that g is a function from A to B and f is a function from B to C. Prove each of these statements. a) If f ◦g is onto, then f must also be onto. b) If f ◦g is one-to-one, then g must also be one-to-one. c) If f ◦g is a bijection,

Step-by-step explanation:

Given that g is a function from A to B and f is a function from B to C.

g: A -->B  and f: B-->C

a) fog= f{g[x)} is a function from A to C

Let fog be onto.  Then we get for any element C in f we got an image in A.

This is possible only if every element of C has an image in B because if not then f cannot be applied to g(x).

Hence proved

b) Let fog be one to one.  

i.e. fog(a) = fog(b) implies a =b

This suggests that f{g(a) }=f{g(b)}

Since f is a function for g(a) and g(b) three exists unique images.

So g(a) = g(b) implies a = b.

g is one to one

c) If fog is one to one and on to then we have f is onto and g is one to one.

Since f is onto every element in g has a preimage so it follows that both f and g are bijective.

x= -3

step-by-step explanation:

step 1: simplify brackets

4 - x - 1 = 6

step 2: simpilfy

3 - x = 6

step 3: subtract 3 from both sides

-x = 6 - 3

step 4: simpilfy 6 - 3

-x = 3

step 5: multiply both sides by -1

x = -3