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