1. **State the problem:** Given the function $$f(x) = a \left(\frac{x-b}{a-b}\right) + b \left(\frac{x-a}{b-a}\right)$$ we want to verify the identity: $$f(a) + f(b) = f(a+b)$$ 2. **Recall the formula and important rules:** The function is a linear combination of terms involving $a$, $b$, and $x$. We will substitute $x = a$, $x = b$, and $x = a+b$ into $f(x)$ and simplify each expression. 3. **Calculate $f(a)$:** $$f(a) = a \left(\frac{a-b}{a-b}\right) + b \left(\frac{a-a}{b-a}\right) = a \cdot 1 + b \cdot 0 = a$$ 4. **Calculate $f(b)$:** $$f(b) = a \left(\frac{b-b}{a-b}\right) + b \left(\frac{b-a}{b-a}\right) = a \cdot 0 + b \cdot 1 = b$$ 5. **Calculate $f(a+b)$:** $$f(a+b) = a \left(\frac{a+b - b}{a-b}\right) + b \left(\frac{a+b - a}{b-a}\right) = a \left(\frac{a}{a-b}\right) + b \left(\frac{b}{b-a}\right)$$ Note that $b - a = -(a - b)$, so: $$f(a+b) = a \frac{a}{a-b} + b \frac{b}{-(a-b)} = \frac{a^2}{a-b} - \frac{b^2}{a-b} = \frac{a^2 - b^2}{a-b}$$ 6. **Simplify $f(a+b)$:** Recall the difference of squares: $$a^2 - b^2 = (a-b)(a+b)$$ So: $$f(a+b) = \frac{(a-b)(a+b)}{a-b} = a + b$$ 7. **Compare $f(a) + f(b)$ and $f(a+b)$:** $$f(a) + f(b) = a + b$$ $$f(a+b) = a + b$$ Thus, $$f(a) + f(b) = f(a+b)$$ **Final answer:** The identity holds true.