Let f be a differentiable function on R and satisfies f x - y - f x - f y - x - y - R- If f -0-2- then the value of f 4 is

F'(x) = lim_h → 0f(x + h) f(x)/h = lim_h → 0f(x) + f(h) f(x)/h ⇒ f'(0) = 2 [∵ f(0) = 0 by putting x = y = 0] ⇒ f(x) = 2x + c Putting x = 0, we get c = 0 ...

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Byju's Answer

Standard XII

Mathematics

Derivative from First Principle

Let f be a di...

Question

Let $f$ be a differentiable function on $R$ and satisfies $f (x + y) = f (x) + f (y) \forall x, y ϵ R$ . If $f^{'} (0) = 2$ , then the value of $f (4)$ is

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Solution

$f^{'} (x) = {lim}_{h \to 0} \frac{f (x + h) - f (x)}{h} = {lim}_{h \to 0} \frac{f (x) + f (h) - f (x)}{h}$

$\Rightarrow f^{'} (0) = 2$ [ $∵ f (0) = 0$ by putting $x = y = 0$ ]

$\Rightarrow f (x) = 2 x + c$

Putting $x = 0,$ we get $c = 0$

$∴ f (x) = 2 x \Rightarrow f (4) = 8$

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Let f be a differentiable function on R and satisfies f(x+y)=f(x)+f(y) ∀ x, y ϵ R. If f′(0)=2, then the value of f(4) is

f′(x)=limh→0f(x+h)−f(x)h=limh→0f(x)+f(h)−f(x)h ⇒f′(0)=2 [∵ f(0)=0 by putting x=y=0] ⇒f(x)=2x+c Putting x=0, we get c=0 ∴f(x)=2x⇒f(4)=8

Let $f$ be a differentiable function on $R$ and satisfies $f (x + y) = f (x) + f (y) \forall x, y ϵ R$ . If $f^{'} (0) = 2$ , then the value of $f (4)$ is

$f^{'} (x) = {lim}_{h \to 0} \frac{f (x + h) - f (x)}{h} = {lim}_{h \to 0} \frac{f (x) + f (h) - f (x)}{h}$

$\Rightarrow f^{'} (0) = 2$ [ $∵ f (0) = 0$ by putting $x = y = 0$ ]

$\Rightarrow f (x) = 2 x + c$

Putting $x = 0,$ we get $c = 0$

$∴ f (x) = 2 x \Rightarrow f (4) = 8$