Let a function f : R → R be given by f(x + y) = f(x) f(y) for all x, y ∈ R and f(x) ≠ 0 for any x ∈ R. If the function f(x) is differentiable at x = 0, show that
f ′ (x) = f ′ (0) f(x) for all x ∈ R. Also, determine f(x).
Text Solution
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Sol. We have,
f(x + y) = f(x) f(y) for all x, y ∈ R
⇒ f(0 + 0) = f(0) f(0) [putting x = y = 0]
⇒ f(0) = f(0) f(0)
⇒ f(0) {1 –f(0)} = 0
⇒ f(0) = 1 [ f(x) ≠ 0 for any x ∈ R]
Since f(x) is differentiable at x = 0. Therefore,
= f ′ (0) exists finitely
Now,
f ′ (x) = 
= 
[By def. of f]
= f(x) 
= f(x) 
[ f(0) = 1]
= f(x). f ′ (0) [using (i)]
Hence, f ′ (x) = f(x). f ′ (0) for all x ∈ R
Now,
f ′ (x) = f(x) f ′ (0)
⇒ 
= f ′ (0)
⇒ log |f(x)| = xf ′ (0) + log A [On integration]
⇒ |f(x)| = Ae xf ′ (0) ⇒ f(x) = ± Ae xf ′ (0) for all x ∈ R
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