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Suppose f:R-IR is a diffeetiable function which satısfies flo) = 0 We wart to show f': IR → IR is a coustant fanction if and cnly if fl*+yl =fx9 + fiy) for all ×, y € IR

Suppose f:R-IR is a diffeetiable function which satısfies flo) = 0 We wart to show f': IR → IR is a coustant fanction if and cnly if fl*+yl =fx9 + fiy) for all ×, y € IR

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Suppose f.R→IR is a diffetinble funtion which satısfies flo) = 0 We want to show f': IR → IR is a coustant function if and anly if flx+yl =frx + fiy) for all ×, y & IR
Transcribed Image Text:Suppose f.R→IR is a diffetinble funtion which satısfies flo) = 0 We want to show f': IR → IR is a coustant function if and anly if flx+yl =frx + fiy) for all ×, y & IR
Expert Solution
Step 1

Given that f:ℝ→ℝ is a differentiable function which satisfies f0=0.

It is required to show that f' : ℝ→ℝ is constant function if and only if fx+y=fx+fy for all x, y∈ℝ.

Suppose if f' : ℝ→ℝ is constant function.

Then it is required to prove that  fx+y=fx+fy for all x, y∈ℝ.

Since f' : ℝ→ℝ is constant function, f'x=a for all x∈ℝ.

Integrating on both sides with respect to x will give

f(x)=ax+b         ...... (1)

where b is the integration constant.

Given that the function satisfies f0=0.

Substituting x=0 in (1) will give

⇒f(0)=a0+b⇒     0=0+b⇒     0=b

Thus the integration constant b=0.

Then (1) becomes f(x)=ax

Now for all x, y∈ℝ,

f(x+y)=ax+y

            =ax+ay

            =fx+fy

Hence proved that if  f' : ℝ→ℝ is constant function, then fx+y=fx+fy for all x, y∈ℝ.        .......... (A)

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