A function f: R -> R satisfies the equation f(x+y) = f(x).f(y) for all x'y E R, f (x) =/= 0.
 
Suppose that the function is differentiable at x=0 and f ' (0) = 2
 
Prove that f '(x) = 2 f ( x)

Asked by ahuja8087 | 7th Jun, 2016, 08:56: PM

Expert Answer:

Q u e s t i o n colon
A space f u n c t i o n space f colon R rightwards arrow R space s a t i s f i e s space t h e space e q u a t i o n space f open parentheses x plus y close parentheses equals f open parentheses x close parentheses f open parentheses y close parentheses space f o r space a l l space x comma y element of R comma space f open parentheses x close parentheses not equal to 0.
S u p p o s e space t h a t space t h e space f u n c t i o n space i s space d i f f e r e n t i a b l e space a t space x equals 0 space a n d space f apostrophe open parentheses 0 close parentheses equals 2.
P r o v e space t h a t space f apostrophe open parentheses x close parentheses equals 2 f open parentheses x close parentheses

S o l u t i o n colon
f apostrophe open parentheses x close parentheses equals stack l i m with x rightwards arrow 0 below fraction numerator f open parentheses x plus h close parentheses minus f open parentheses x close parentheses over denominator h end fraction
space space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator f open parentheses x close parentheses f open parentheses h close parentheses minus f open parentheses x close parentheses over denominator h end fraction
space space space space space space space space equals stack l i m f open parentheses x close parentheses with x rightwards arrow 0 below open square brackets fraction numerator f open parentheses h close parentheses minus 1 over denominator h end fraction close square brackets
space space space space space space space space equals stack l i m f open parentheses x close parentheses with x rightwards arrow 0 below open square brackets fraction numerator f open parentheses h close parentheses minus f open parentheses 0 close parentheses over denominator h end fraction close square brackets space space space space space space space space space open square brackets because f open parentheses x plus 0 close parentheses equals f open parentheses x close parentheses cross times f open parentheses 0 close parentheses rightwards double arrow f open parentheses 0 close parentheses equals 1 close square brackets
rightwards double arrow f apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses space f apostrophe open parentheses 0 close parentheses
rightwards double arrow f apostrophe open parentheses x close parentheses equals 2 f open parentheses x close parentheses space space space space space space space space space space space space open square brackets because space f apostrophe open parentheses 0 close parentheses equals 2 close square brackets

Answered by Vimala | 8th Jun, 2016, 11:19: AM

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