Let f(x + y) = f(x)f(y) ∀ x, y ∈R. Suppose that f(3) = 3 and f' (0) = 11 then f'(3) =

f(x+y)=f(x) f(y) f(0)=1 f(3)=3, f1(0)=11 f1(x)=lth→0 f(x+h)-f(x)h =lth→0 f(x) f(h)-f(x)h =lth→0 f(x) f(h)-f(0)h =f(x) f1(0) f1(3)=f(3) f1(0) =3(11)=33

Let f(x + y) = f(x)f(y) ∀ x, y ∈R. Suppose that f(3) = 3 and f' (0) = 11 then f'(3) =

f(x+y)=f(x) f(y) f(0)=1 f(3)=3, f1(0)=11 f1(x)=lth→0 f(x+h)-f(x)h =lth→0 f(x) f(h)-f(x)h =lth→0 f(x) f(h)-f(0)h =f(x) f1(0) f1(3)=f(3) f1(0) =3(11)=33

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Let $f (x + y) = f (x) f (y) \forall x, y \in R .$ Suppose that $f (3) = 3$ and $f' (0) = 11$ then $f' (3) =$

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Detailed Solution

$f (x + y) = f (x) f (y) f (0) = 1 f (3) = 3, f^{1} (0) = 11 f^{1} (x) = \underset{h \to 0}{lt} \frac{f (x + h) - f (x)}{h} = \underset{h \to 0}{lt} \frac{f (x) f (h) - f (x)}{h} = \underset{h \to 0}{lt} f (x) \frac{f (h) - f (0)}{h} = f (x) f^{1} (0) f^{1} (3) = f (3) f^{1} (0) = 3 (11) = 33$