# Functions – Problems

### Functions – Problems

1. f: R → R defined by f (x) = sinx and g: R → R defined by g (x) = x², fog (x) =?

Solution: Given that

f (x) = sin x

g (x) = x²

fog (x) = f [g (x)]

= f [x²]

= sin x².

2. If f: [-6, 6] → R is defined by f (x) = x² – 3 for x ϵ R then (fofof) (-1) + (fofof) (0) + (fofof) (1) =?

Solution: Given

f (x) = x² – 3

Now f (-1) = (-1)² – 3 = -2

fof (-1) = f (-2) = (-2)² – 3 = 1

⇒ fofof (-1) = f (1) = (1)² – 3 = -2

Now f (0) = 0 – 3 = -3

fof (0) = (-3)² – 3

= 9 – 3

= 6

fofof (0) = (6)² – 3

= 36 – 3

= 33

Now f (1) = (1)² – 3

= – 2

fof (1) = (-2)² – 3

= 4 – 3

= 1

fofof (1) = (1)² – 3

= -2

fofof (-1) + fofof (1) + fofof (-1) = -2 + 3 – 2 = 29

f (4√2) = (4√2)² – 3

= 32 – 3

= 29.

3. If y = f (x) = $$\frac{2x-1}{x-2}$$ then f (y)

Solution: y = f (x) = $$\frac{2x-1}{x-2}$$,

x = f (y) = $$\frac{2y-1}{y-2}$$ … 1

$$f(x)=\frac{2x-1}{x-2}$$.

$$y=\frac{2x-1}{x-2}$$.

yx – 2y = 2x – 1

yx – 2x = – (1 – 2y)

x (y – 2) = 2y – 1

$$x=\frac{2y-1}{y-2}$$,

∴ f (y) = x = $$\frac{2y-1}{y-2}$$ (since from equation 1).

4. If $$f(x)=\frac{2x+1}{x+2}$$ then what is the inverse function.

Solution: $$f(x)=\frac{(2x+1)}{(x+2)}$$,

Let$$y=\frac{(2x+1)}{(x+2)}$$,

Multiply by (x + 2)

⇒ y (x + 2) = 2x + 1

⇒ yx + 2y = 2x + 1

⇒ yx – 2x = 1 – 2y

⇒ x (y – 2) = 1 – 2y

⇒ $$x=\frac{1-2y}{(y-2)}$$,

Then the inverse is $${{f}^{-1}}(x)=\frac{(1-2x)}{(x-2)}$$.