Given:
Consider the given function:
[tex]f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}[/tex]
To prove:
[tex]f(a)+f(b)=f(a+b)[/tex]
Solution:
We have,
[tex]f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}[/tex]
Substituting [tex]x=a[/tex], we get
[tex]f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}[/tex]
[tex]f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}[/tex]
[tex]f(a)=0+a[/tex]
[tex]f(a)=a[/tex]
Substituting [tex]x=b[/tex], we get
[tex]f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}[/tex]
[tex]f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}[/tex]
[tex]f(b)=b+0[/tex]
[tex]f(b)=b[/tex]
Substituting [tex]x=a+b[/tex], we get
[tex]f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}[/tex]
[tex]f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}[/tex]
[tex]f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}[/tex]
[tex]f(a+b)=\dfrac{b^2-a^2}{b-a}[/tex]
Using the algebraic formula, we get
[tex]f(a+b)=\dfrac{(b-a)(b+a)}{b-a}[/tex] [tex][\because b^2-a^2=(b-a)(b+a)][/tex]
[tex]f(a+b)=b+a[/tex]
[tex]f(a+b)=a+b[/tex] [Commutative property of addition]
Now,
[tex]LHS=f(a)+f(b)[/tex]
[tex]LHS=a+b[/tex]
[tex]LHS=f(a+b)[/tex]
[tex]LHS=RHS[/tex]
Hence proved.
