Question

If f(x + y) = f(x) f(y) for all x, y ∈ R and f(5) = 2, f′(0) = 3, then using the definition of derivatives, find f′(5).

Answer 1

f'(x) = `lim_(h -> 0) (f(x + h) - f(x))/h`

Putting x = 5

f'(5) = `lim_(h -> 0) (f(5 + h) - f(5))/h`

Using f(x + y) = f(x) f(y)

f'(5) = `lim_(h -> 0) (f(5) xx f(h) - f(5))/h`

f'(5) = `lim_(h -> 0) (f(5) [f(h) - 1])/h`

Putting f(5) = 2

f′(5) = `lim_(h -> 0) (2 xx [f(h) - 1])/h`

f′(5) = `2 lim_(h -> 0) ([f(h) - 1])/h`   ....(1)

Now, since f'(x) = `lim_(h -> 0) (f(x + h) - f(x))/h`

Putting x = 0

f′(0) = `lim_(h -> 0) (f(0 + h) - f(0))/h`

f′(0) = `lim_(h -> 0) (f(h) - f(0))/h`

Putting f'(0) = 3

3 = `lim_(h -> 0) (f(h) - f(0))/h`

`lim_(h -> 0) (f(h) - f(0))/h` = 3  ....(2)

Finding the value of f(0):

Given that f(x + y) = f(x) f(y) and f(5) = 2

Putting x = 0, y = 5

f(0 + 5) = f(0) f(5)

f(5) = f(0) f(5)

Putting f(5) = 2

2 = F(0) × 2

1 = f(0)

f(0) = 1

Putting f(0) = 1 in equation (2)

`lim_(h -> 0) (f(h) - f(0))/h` = 3

`lim_(h -> 0) (f(h) - 1)/h` = 3    ....(3)

Putting equation (3) in equation (1)

f′(5) = `2 lim_(h -> 0) ([f(h) - 1])/h`

f′(5) = 2 × 3

f′(5) = 6