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Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as
f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.
Let f, g and h be functions from R to R. Show that
To prove:
Find gof and fog, if
(i)
(ii)
(i)
(ii)
If, show that f o f(x) = x, for all. What is the inverse of f?
It is given that.
Hence, the given function f is invertible and the inverse of f is f itself.
State with reason whether following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
(i) f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10
∴f is not one-one.
Hence, function f does not have an inverse.
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.
∴g is not one-one,
Hence, function g does not have an inverse.
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
∴Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.
Show that f: [−1, 1] → R, given byis one-one. Find the inverse of the function f: [−1, 1] → Range f.
(Hint: For y ∈Range f, y =, for some x in [−1, 1], i.e.,)
f: [−1, 1] → R is given as
Let f(x) = f(y).
∴ f is a one-one function.
It is clear that f: [−1, 1] → Range f is onto.
∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f: [−1, 1] → Range f exists.
Let g: Range f → [−1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f: [−1, 1] → Range f is onto, we have:
Now, let us define g: Range f → [−1, 1] as
∴gof = and fog =
f−1 = g
⇒
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
f: R → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y).
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3.
Therefore, for any y ∈ R, there exists such that
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: R→ R by.
∴
Hence, f is invertible and the inverse of f is given by
Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by, where R+ is the set of all non-negative real numbers.
f: R+ → [4, ∞) is given as f(x) = x2 + 4.
One-one:
Let f(x) = f(y).
∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x2 + 4.
Therefore, for any y ∈ R, there exists such that
.
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: [4, ∞) → R+ by,
∴
Hence, f is invertible and the inverse of f is given by
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with.
f: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6x − 5.
∴f is onto, thereby range f = [−5, ∞).
Let us define g: [−5, ∞) → R+ as
We now have:
∴and
Hence, f is invertible and the inverse of f is given by
Let f: X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,
fog1(y) = IY(y) = fog2(y). Use one-one ness of f).
Let f: X → Y be an invertible function.
Also, suppose f has two inverses (say).
Then, for all y ∈Y, we have:
Hence, f has a unique inverse.
Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.
Function f: {1, 2, 3} → {a, b, c} is given by,
f(1) = a, f(2) = b, and f(3) = c
If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:
∴and, where X = {1, 2, 3} and Y= {a, b, c}.
Thus, the inverse of f exists and f−1= g.
∴f−1: {a, b, c} → {1, 2, 3} is given by,
f−1(a) = 1, f−1(b) = 2, f-1(c) = 3
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h: {1, 2, 3} → {a, b, c} as
h(1) = a, h(2) = b, h(3) = c, then we have:
∴, where X = {1, 2, 3} and Y = {a, b, c}.
Thus, the inverse of g exists and g−1 = h ⇒ (f−1)−1 = h.
It can be noted that h = f.
Hence, (f−1)−1 = f.
Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,
(f−1)−1 = f.
Let f: X → Y be an invertible function.
Then, there exists a function g: Y → X such that gof = IXand fog = IY.
Here, f−1 = g.
Now, gof = IXand fog = IY
⇒ f−1of = IXand fof−1= IY
Hence, f−1: Y → X is invertible and f is the inverse of f−1
i.e., (f−1)−1 = f.
If f: R → R be given by, then fof(x) is
(A) (B) x3 (C) x (D) (3 − x3)
f: R → R is given as.
The correct answer is C.
Letbe a function defined as. The inverse of f is map g: Range
(A) (B)
(C) (D)
It is given that
Let y be an arbitrary element of Range f.
Then, there exists x ∈such that
Let us define g: Rangeas
Now,
∴
Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map g: Range, which is given by
The correct answer is B.
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