Let ϕ(x)=(x−b)(x−c)(a−b)(a−c)f(a)+(x−c)(x−a)(b−c)(b−a)f(b)+(x−a)(x−b)(c−a)(c−b)f(c)−f(x) Where a < c < b and f11(x) exists at all points in (a,b) . Then, there exists a real number μ a < μ < b such that f(a)(a−b)(a−c)+f(b)(b−c)(b−a)+f(c)(c−a)(c−b)=

Let ϕ(x)=(x−b)(x−c)(a−b)(a−c)f(a)+(x−c)(x−a)(b−c)(b−a

1.	Let ϕ(x)=(x−b)(x−c)(a−b)(a−c)f(a)+(x−c)(x−a)(b−c)(b−a)f(b)+(x−a)(x−b)(c−a)(c−b)f(c)−f(x) Where a < c < b and f11(x) exists at all points in (a,b) . Then, there exists a real number μ a < μ < b such that f(a)(a−b)(a−c)+f(b)(b−c)(b−a)+f(c)(c−a)(c−b)=
Answer» Let $ϕ (x) = \frac{(x - b) (x - c)}{(a - b) (a - c)} f (a) + \frac{(x - c) (x - a)}{(b - c) (b - a)} f (b) + \frac{(x - a) (x - b)}{(c - a) (c - b)} f (c) - f (x)$ Where a < c < b and $f^{11} (x)$ exists at all points in (a,b) . Then, there exists a real number $μ$ a < $μ$ < b such that $\frac{f (a)}{(a - b) (a - c)} + \frac{f (b)}{(b - c) (b - a)} + \frac{f (c)}{(c - a) (c - b)} =$