Functions?

Question

Can u help me state whether each of the following functions is:-
(assuming that the domain and codomain are both contained in R (real numbers) )

(i) one to one (ii) onto (iii) both (iv) neither

(a) y = 1-x^2

(b) y = -3x

(c) y = x(1-x)(1+x)


(d) y = Ln(x)

(e) y = Sin(x)

(f) y = Tan(x)

(g) y = e^x

(h) y = 10x^3+3

Answer 1

I'll do some in detail and provide the answers for the rest, you really should be able to fill in the missing logic.
(a) y = 1 - x^2: not one-to-one since y(-1) = y(1) = 0. Also not onto since there is no value of x for which y > 1.

(b) both 1-1 and onto. You should be able to prove this easily.

(c) Should be obvious that this is not one-to-one (hint: you can read off three values of x for which y = 0). This is onto - note that it's continuous and has limit -infinity as x tends to +infinity, and limit +infinity as x tends to -infinity; thus y must pass through every real number (formal definition of the limit and intermediate value theorem come into play here, but really it's obvious).

(d) One-to-one and onto, but note only defined on the domain (0, infinity).

(e) Neither one-to-one nor onto; this should be obvious from the graph.

(f) Onto but not one-to-one; also obvious from the graph. Note that no periodic function can be one-to-one!

(g) One-to-one but not onto (e^x > 0 for all x) - note this is the inverse function of (d). The fact that this function is always > 0 directly relates to the domain of (d) being only for x>0.

(h) Both one-to-one and onto. Easy to prove based on the fact that x^3 is 1-1 and onto.

Answer 2

Although it may not apply in this particular case, there are plenty of operations in R that are not one-to-one (arcsin, square root...); there you have to create cyclical groups and talk of "congruence". For example, number the days of the week 0=sunday, 1=monday...
How many days from sunday to wednesday? answer can be 3, 10, 17, 24... In base 7, all these numbers are congruent to 3 (i.e., they all have a remainder of 3 when divided by 7).

However, here we have simple cases. It may appear trivial, but you have to practice checking whether a function is one-to-one and onto. Later, you may have to check trickier functions.

one-to-one: for any value of x, y will have one, and only one value.
For example, y = -3x, if x = -2 then y can only be +6. y cannot have another value.

The ill-defined square root can have more than one answer: y = sqrt(x) if x = 4 then y = 2 and y = -2 are both possible values. To be a proper function, the square root must be "well-defined" (usually as the absolute value -- i.e., positive answers only); only then is it one-to-one.

onto: all values of the domain are used: for any value of y, you can find a value of x that can give you the desired y.

in the case of y = 1 - x^2, no matter how hard you try, you cannot find an x such that y = 2 (because x^2 is always non-negative, then y can never be more than 1).

If you need that function to be "onto", then you define the range as "all the real numbers such that y =< 1"

onto is not to be confused with "surjective" meaning that for any desired value y, there can only be 1 value of x. In this case again, x=1 gives y=0 AND x=-1 gives y=0; not surjective.

Answer 3

all functions here are 1-to-1 since most of functions in real numbers
are 1-to-1 functions

Answer 4

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Answer 5

a - iv
b - iii
c - i or iv (Not sure)
d - iv
e - iv
f - i
g - i
h - iii ( I guess )