Advertisements
Advertisements
Question
Check if the following function has an inverse function. If yes, find the inverse function.
f(x) = `(6x - 7)/3`
Advertisements
Solution
f(x) = `(6x - 7)/3`
Let f(x1) = f(x2)
∴ `(6x_1 - 7)/3 = (6x_2 - 7)/3`
∴ x1 = x2
∴ f is a one-one function
f(x) = `(6x - 7)/3` = y (say)
∴ x = `(3"y" + 7)/6`
∴ For every y, we can get x
∴ f is an onto function.
∴ x = `(3y + 7)/6`
= f–1 (y)
Replacing y by x, we get
∴ f–1 (x) = `(3x + 7)/6`
APPEARS IN
RELATED QUESTIONS
Let f : {2, 4, 5} → {2, 3, 6} and g : {2, 3, 6} → {2, 4} be given by f = {(2, 3), (4, 6), (5, 2)} and g = {(2, 4), (3, 4), (6, 2)}. Write down g ° f
If f(x) = 2x2 + 3, g (x) = 5x − 2, then find f ° g
If f(x) = 2x2 + 3, g (x) = 5x − 2, then find g ° g
Verify that f and g are inverse functions of each other, where f(x) = `(x - 7)/4`, g(x) = 4x + 7
Check if the following function has an inverse function. If yes, find the inverse function.
f(x) = 5x2
Check if the following function has an inverse function. If yes, find the inverse function.
f(x) = `sqrt(4x + 5)`
Check if the following function has an inverse function. If yes, find the inverse function.
f(x) = 9x3 + 8
Check if the following function has an inverse function. If yes, find the inverse function.
f(x) = `{(x + 7, x < 0),(8 - x, x ≥ 0):}`
If f(x) = `{(x^2 + 3, x ≤ 2),(5x + 7, x > 2):}`, then find f(2)
If f(x) = 2{x} + 5x, where {x} is fractional part function of x, then find f(– 1)
If f(x) = 2{x} + 5x, where {x} is fractional part function of x, then find `"f"(1/4)`
Solve the following for x, where |x| is modulus function, [x] is greatest integer function, [x] is a fractional part function.
|x + 4| ≥ 5
Solve the following for x, where |x| is modulus function, [x] is greatest integer function, [x] is a fractional part function.
2|x| = 5
Solve the following for x, where |x| is modulus function, [x] is greatest integer function, [x] is a fractional part function.
{x} > 4
Solve the following for x, where |x| is modulus function, [x] is greatest integer function, [x] is a fractional part function.
{x} = 0
Solve the following for x, where |x| is modulus function, [x] is greatest integer function, [x] is a fractional part function.
2{x} = x + [x]
Answer the following:
Find whether the following function is onto or not.
f : Z → Z defined by f(x) = 6x – 7 for all x ∈ Z
Answer the following:
Find whether the following function is onto or not.
f : R → R defined by f(x) = x2 + 3 for all x ∈ R
Answer the following:
Find f ° g and g ° f: f(x) = 3x – 2, g(x) = x2
Answer the following:
Find f ° g and g ° f: f(x) = 256x4, g(x) = `sqrt(x)`
Answer the following:
If f(x) = `(2x - 1)/(5x - 2), x ≠ 5/2` show that (f ° f) (x) = x
Answer the following:
Solve the following for x, where |x| is modulus function, [x] is greatest interger function, {x} is a fractional part function
2[2x − 5] − 1 = 7
Answer the following:
Solve the following for x, where |x| is modulus function, [x] is greatest interger function, {x} is a fractional part function
[x2] − 5[x] + 6 = 0
Answer the following:
Solve the following for x, where |x| is modulus function, [x] is greatest interger function, {x} is a fractional part function
`[x/2] + [x/3] = (5x)/6`
Answer the following:
Find the domain of the following function.
f(x) = `sqrt(1 - sqrt(1 - sqrt(1 - x^2)`
Answer the following:
Find (f ° g) (x) and (g ° f) (x)
f(x) = ex, g(x) = log x
Answer the following:
Find (f ° f) (x) if f(x) = `x/sqrt(1 + x^2)`
Let F(x) = ex, G(x) = e-x and H(x) = G[F(x)], where x is a real variable. Then `"dH"/"dx"`at x = 0 is ______.
If f(x) = `sin^2x + sin^2(x + pi/3) + cosx cos(x + pi/3) and g(5/4) = 1`, then (gof)(x) is equal to: ______
If f(x) =bx - 7 and f(-1) = 4, then b = ______.
The inverse of f(x) = `2/3 (10^x - 10^-x)/(10^x + 10^-x)` is ______.
lf f : [1, ∞) `rightarrow` [2, ∞) is given by f(x) = `x + 1/x`, then f–1(x) is equal to ______.
The value of `int_-1^3 (|x - 2| + [x]) dx` is equal to ______.
(where [.] denotes greatest integer function)
Given, the function f(x) = `(a^x + a^(-x))/2 (a > 2)`, then f(x + y) + f (x − y) is equal to ______.
