Answer to Question #346724 in Real Analysis for Dhruv rawat

Question

Let ϕ and Ψ be function defined on [-3,5], such that both are continuous on [-3,5], derivable in [-3,5] and ϕ'(x)= Ψ'(x) ∀ x∈]-3,5[. Prove that

ϕ(x)= Ψ(x) +c ∀ x∈ [-3,5] , where c is a real constant

Answer 1

PROOF

Denote $h(x)=\Phi (x)-\Psi(x)$ . The function $\Phi$ and $\Psi$ are continuous and differentiable on the segment $[-3,5]$ , therefore $h$ is also differentiable on the $[-3,5]$ .$h'(x)=\Phi'(x)-\Psi'(x)=0$ for all $x\in$$]-3,5[.$ Let $x_{1}, x_{2}\in ]-3,5[$ and $x_{1}<x_{2}$ .

By the Mean Value Theorem for the function $h$ there exists a point $d\in ]x_{1},x_{2}[$ , such that

$h(x_{1})-h(x_{2})=h'(d)\cdot(x_{1}-x_{2})=0$ .

Hence, $h(x_{1})-h(x_{2})=0 \Rightarrow h(x_{1})=h(x_{2})$ . Since $x_{1}, x_{2}$ were any two values of $x\in]-3,5[$ this is what it means for a function $h$ to be constant on the interval $]-3,5[.$ So, $h(x)=c$ for all $x\in]-3,5[$ , where $c\in\R$ .

$\lim_{x\rightarrow (-3)^{+}}h(x)= \lim_{x\rightarrow 5^{-}}h(x)=c$ . Since $h$ is continuous at the points $x=-3$ and $x=5$ , then $h(-3)=h(5)=c$ . Therefore,

$h(x)=c$ for all $x\in[-3,5]$ $\Rightarrow$ $\Phi(x)=\Psi(x)+c$ for all $x\in[-3,5]$ ,$c\in\R$ .