Questions on Function and Inverse Function

Functions and Inverse Functions – 10 Practice Questions

Question 1. Determine whether the relation

R = {(1,2), (2,4), (3,6), (4,8)}

defines a function.

Show Solution

A relation is a function if every input has exactly one output.

Here the inputs are 1, 2, 3, and 4, and each input has only one corresponding output.

Therefore, R is a function.

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Question 2. Given

f(x) = 3x - 5

find f(2), f(-1), and f(0).

Show Solution

For x = 2:

f(2) = 3(2) - 5 = 6 - 5 = 1

For x = -1:

f(-1) = 3(-1) - 5 = -3 - 5 = -8

For x = 0:

f(0) = 3(0) - 5 = -5

Hence, f(2) = 1, f(-1) = -8, and f(0) = -5.

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Question 3. If

f(x) = x² + 2x - 3

find f(a + 1).

Show Solution

Replace x by a + 1:

f(a + 1) = (a + 1)² + 2(a + 1) - 3

Expanding,

= a² + 2a + 1 + 2a + 2 - 3

= a² + 4a

Therefore, f(a + 1) = a² + 4a.

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Question 4. Find the domain of

f(x) = (2x + 1)/(x - 4)

Show Solution

The denominator cannot be zero.

x - 4 ≠ 0

Therefore,

x ≠ 4

Hence the domain is all real numbers except 4.

Domain: (-∞, 4) ∪ (4, ∞)

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Question 5. Let

f(x) = 2x + 3,    g(x) = x²

Find (f ∘ g)(x) and (g ∘ f)(x).

Show Solution

First,

(f ∘ g)(x) = f(g(x))

Since g(x) = x²,

f(g(x)) = f(x²) = 2x² + 3

Therefore,

(f ∘ g)(x) = 2x² + 3

Now,

(g ∘ f)(x) = g(f(x))

= (2x + 3)²

= 4x² + 12x + 9

Therefore,

(g ∘ f)(x) = 4x² + 12x + 9

Hence, f ∘ g is generally not equal to g ∘ f.

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Question 6. Find the inverse of

f(x) = 5x - 7

Show Solution

Let

y = 5x - 7

Interchange x and y:

x = 5y - 7

Now solve for y:

x + 7 = 5y

y = (x + 7)/5

Therefore,

f^-1(x) = (x + 7)/5

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Question 7. Find the inverse of

f(x) = (x - 2)/3

Show Solution

Let

y = (x - 2)/3

Interchange x and y:

x = (y - 2)/3

Multiply both sides by 3:

3x = y - 2

Hence,

y = 3x + 2

Therefore,

f^-1(x) = 3x + 2

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Question 8. Find the inverse of

f(x) = (2x + 3)/(x - 1),   x ≠ 1

Show Solution

Let

y = (2x + 3)/(x - 1)

Multiply both sides by x - 1:

y(x - 1) = 2x + 3

Expanding,

yx - y = 2x + 3

Collect the x terms:

yx - 2x = y + 3

Factor:

x(y - 2) = y + 3

Therefore,

x = (y + 3)/(y - 2)

Replacing y by x:

f^-1(x) = (x + 3)/(x - 2)

The inverse is defined for x ≠ 2.

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Question 9. Does the function

f(x) = x²

have an inverse on the set of all real numbers? If not, restrict its domain and find its inverse.

Show Solution

The function f(x) = x² is not one-to-one on the real numbers.

For example,

f(2) = 4

and

f(-2) = 4

Two different inputs give the same output, so the function does not have an inverse on all real numbers.

Restrict the domain to x ≥ 0.

Now let

y = x²

Interchanging x and y gives

x = y²

Therefore,

y = √x

Hence, for x ≥ 0,

f^-1(x) = √x

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Question 10. Let

f(x) = (3x - 2)/(x + 4)

Find f^-1(x) and verify that f(f^-1(x)) = x.

Show Solution

Let

y = (3x - 2)/(x + 4)

Multiply both sides by x + 4:

y(x + 4) = 3x - 2

Expanding,

yx + 4y = 3x - 2

Collect x terms:

yx - 3x = -2 - 4y

Factor x:

x(y - 3) = -2 - 4y

Thus,

x = (4y + 2)/(3 - y)

Replacing y by x:

f^-1(x) = (4x + 2)/(3 - x)

Verification:

f(f^-1(x)) = [3((4x + 2)/(3 - x)) - 2] / [((4x + 2)/(3 - x)) + 4]

The numerator simplifies to

14x/(3 - x)

and the denominator simplifies to

14/(3 - x)

Therefore,

f(f^-1(x)) = x

Hence the inverse is verified.

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Important Note:
f^-1(x) does not mean 1/f(x).
f^-1(x) represents the inverse function.

Question 11. Determine whether the function

f(x) = x³ - 4x + 1

is one-to-one on R.

Show Solution

For a function to have an inverse on R, it must be one-to-one.

Differentiate:

f'(x) = 3x² - 4

The derivative is zero when

3x² - 4 = 0

so

x = ± 2/√3.

The derivative changes sign, so the function is not strictly increasing or strictly decreasing on all of R.

Therefore, f is not one-to-one on R and hence does not have an inverse on the whole real line.

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Question 12. Let

f(x) = e^2x+1.

Find f^-1(x) and state its domain.

Show Solution

Let

y = e^2x+1.

Take natural logarithm on both sides:

ln y = 2x + 1.

Therefore,

2x = ln y - 1

and

x = (ln y - 1)/2.

Interchanging x and y gives

f^-1(x) = (ln x - 1)/2.

Since the logarithm requires x > 0, the domain of the inverse is

(0, ∞).

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Question 13. Find the inverse of

f(x) = ln(3x - 2)

and determine the domain and range of both f and f^-1.

Show Solution

First, for f(x) to be defined,

3x - 2 > 0.

Thus,

x > 2/3.

Therefore,

Domain of f: (2/3, ∞)

Range of f: R

Now let

y = ln(3x - 2).

Exponentiating both sides gives

e^y = 3x - 2.

Hence,

x = (e^y + 2)/3.

Therefore,

f^-1(x) = (e^x + 2)/3.

Domain of f^-1: R

Range of f^-1: (2/3, ∞)

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Question 14. Let

f(x) = x/(1 + x),   x ≠ -1.

Find the inverse function and determine whether f is self-inverse.

Show Solution

Let

y = x/(1 + x).

Then

y(1 + x) = x.

Expanding,

y + xy = x.

Hence,

y = x - xy = x(1 - y).

Therefore,

x = y/(1 - y).

Thus,

f^-1(x) = x/(1 - x).

Since

f(x) = x/(1 + x)

and

f^-1(x) = x/(1 - x),

we have

f^-1 ≠ f.

Therefore, f is not self-inverse.

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Question 15. Suppose

f(x) = x² + 4x + 7,   x ≥ -2.

Find f^-1(x).

Show Solution

Complete the square:

f(x) = x² + 4x + 7

= (x + 2)² + 3.

Let

y = (x + 2)² + 3.

Then

y - 3 = (x + 2)².

Since x ≥ -2, we have x + 2 ≥ 0, so we take the positive square root:

x + 2 = √(y - 3).

Thus,

x = -2 + √(y - 3).

Therefore,

f^-1(x) = -2 + √(x - 3).

The domain of the inverse is x ≥ 3.

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Question 16. Let

f(x) = 2x - 1,

and

g(x) = (x + 1)/2.

Show that f and g are inverse functions.

Show Solution

Two functions are inverses if both compositions give the identity function.

First,

f(g(x)) = 2[(x + 1)/2] - 1.

Simplifying,

f(g(x)) = x + 1 - 1 = x.

Now,

g(f(x)) = [(2x - 1) + 1]/2.

Thus,

g(f(x)) = 2x/2 = x.

Therefore,

f(g(x)) = g(f(x)) = x.

Hence, f and g are inverse functions.

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Question 17. If

f(x) = (ax + b)/(cx + d),

where ad - bc ≠ 0, find the general formula for f^-1(x).

Show Solution

Let

y = (ax + b)/(cx + d).

Cross multiply:

y(cx + d) = ax + b.

Expand:

cxy + dy = ax + b.

Collect the terms involving x:

cxy - ax = b - dy.

Factor x:

x(cy - a) = b - dy.

Thus,

x = (b - dy)/(cy - a).

Multiplying numerator and denominator by -1:

x = (dy - b)/(a - cy).

Therefore,

f^-1(x) = (dx - b)/(a - cx).

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Question 18. Let

f(x) = x + 1/x,   x > 1.

Find the inverse function.

Show Solution

Let

y = x + 1/x.

Multiply by x:

yx = x² + 1.

Rearranging,

x² - yx + 1 = 0.

Using the quadratic formula,

x = [y ± √(y² - 4)]/2.

Since the original domain is x > 1, we must choose the positive branch:

x = [y + √(y² - 4)]/2.

Hence,

f^-1(x) = [x + √(x² - 4)]/2.

For x > 1, the minimum value approached is 2, so the inverse is defined for

x > 2.

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Question 19. Prove that if f and g are invertible functions, then

(f ∘ g)^-1 = g^-1 ∘ f^-1.

Show Solution

Let

h = f ∘ g.

We want to show that

h^-1 = g^-1 ∘ f^-1.

Consider the composition

(f ∘ g) ∘ (g^-1 ∘ f^-1).

Using associativity,

= f ∘ (g ∘ g^-1) ∘ f^-1.

Since

g ∘ g^-1 = I,

we get

= f ∘ I ∘ f^-1

= f ∘ f^-1

= I.

Similarly,

(g^-1 ∘ f^-1) ∘ (f ∘ g) = I.

Therefore,

(f ∘ g)^-1 = g^-1 ∘ f^-1.

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Question 20. Let

f(x) = x⁵ + x + 1.

Show that f has an inverse on R, even though an explicit elementary formula for the inverse is not required.

Show Solution

Differentiate:

f'(x) = 5x⁴ + 1.

Since

5x⁴ + 1 > 0

for every real x, the function is strictly increasing on R.

Therefore, f is one-to-one.

Also,

lim_x→∞ f(x) = ∞

and

lim_x→-∞ f(x) = -∞.

Thus the range of f is all of R.

Hence,

f : R → R

is bijective and therefore possesses an inverse function

f^-1 : R → R.

An elementary closed-form expression for f^-1 is not needed to prove that the inverse exists.

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University-Level Key Ideas

A function has an inverse if it is one-to-one on its domain.
A strictly increasing or strictly decreasing function is one-to-one.
For differentiable functions, the sign of f'(x) is often useful for proving invertibility.
The domain of f^-1 is the range of f, and the range of f^-1 is the domain of f.