Advertisements
Advertisements
Question
Let `f : R {(-1)/3} → R - {0}` be defined as `f(x) = 5/(3x + 1)` is invertible. Find f–1(x).
Advertisements
Solution
Given, `f(x) = 5/(3x + 1)` and is invertible.
So, we must check for invertibility.
Now, let f(x) = y = `5/(3x + 1)`
`\implies` y(3x + 1) = 5
`\implies` 3xy + y = 5
`\implies` 3xy = 5 – y
`\implies x = (5 - y)/(3y)`
∴ `f^-1(y) = (5 - y)/(3y)`
Now put y = x
`\implies f^-1(x) = (5 - x)/(3x)`
APPEARS IN
RELATED QUESTIONS
if f(x) = `(4x + 3)/(6x - 4), x ≠ 2/3` show that fof(x) = x, for all x ≠ 2/3 . What is the inverse of f?
State with reason whether following functions have inverse g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Show that f: [−1, 1] → R, given by f(x) = `x/(x + 2)` is one-one. Find the inverse of the function f: [−1, 1] → Range f.
(Hint: For y in Range f, y = `f(x) = x/(x +2)` for some x in [-1, 1] ie x = `2y/(1-y)`
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with `f^(-1)(y) = ((sqrt(y +6) - 1)/3)`
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. Show that the inverse of f−1 is f, i.e., (f−1)−1 = f.
If f: R → R be given by `f(x) = (3 - x^3)^(1/3)` , then fof(x) is
(A) `1/(x^3)`
(B) x3
(C) x
(D) (3 − x3)
Let `f:R - {-4/3} -> R` be a function defined as `f(x) = (4x)/(3x + 4)`. The inverse of f is map g Range `f -> R -{- 4/3}`
(A) `g(y) = (3y)/(3-4y)`
(B) `g(y) = (4y)/(4 - 3y)`
(C) `g(y) = (4y)/(3 - 4y)`
(D) `g(y) = (3y)/(4 - 3y)`
Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by g (x) = αx + β, then what value should be assigned to α and β
Let f: R → R be defined by f(x) = 3x 2 – 5 and g: R → R by g(x) = `x/(x^2 + 1)` Then gof is ______.
Let f: A → B and g: B → C be the bijective functions. Then (g o f)–1 is ______.
If f(x) = (ax2 + b)3, then the function g such that f(g(x)) = g(f(x)) is given by ____________.
Let f : N → R : f(x) = `((2"x"−1))/2` and g : Q → R : g(x) = x + 2 be two functions. Then, (gof) `(3/2)` is ____________.
Let f : R → R be the functions defined by f(x) = x3 + 5. Then f-1(x) is ____________.
If f(x) = (ax2 – b)3, then the function g such that f{g(x)} = g{f(x)} is given by ____________.
Which one of the following functions is not invertible?
If f : R → R defind by f(x) = `(2"x" - 7)/4` is an invertible function, then find f-1.
If f : R → R defined by f(x) `= (3"x" + 5)/2` is an invertible function, then find f-1.
`f : x -> sqrt((3x^2 - 1)` and `g : x -> sin (x)` then `gof : x ->`?
Domain of the function defined by `f(x) = 1/sqrt(sin^2 - x) log_10 (cos^-1 x)` is:-
If `f(x) = 1/(x - 1)`, `g(x) = 1/((x + 1)(x - 1))`, then the number of integers which are not in domian of gof(x) are
Let A = `{3/5}` and B = `{7/5}` Let f: A → B: f(x) = `(7x + 4)/(5x - 3)` and g:B → A: g(y) = `(3y + 4)/(5y - 7)` then (gof) is equal to
Let 'D' be the domain of the real value function on Ir defined by f(x) = `sqrt(25 - x^2)` the D is :-
If f: A → B and G B → C are one – one, then g of A → C is
If f(x) = [4 – (x – 7)3]1/5 is a real invertible function, then find f–1(x).
