Lagrange's mean value theorem

$\frac{f(b)-f(a)}{b-a}=f^{\prime }(c)$

• Two real numbers $a<b$
• Function $f(x)$.
  • Continuous on closed interval $[a,b$]
  • Differentiable on open interval [$ (a, b)
• In this case, there exists a point $c$ on the open interval $(a,b)$ and the following holds
• $\frac{f(b)-f(a)}{b-a}=f^{\prime }(c)$
• This is called Lagrange’s mean value theorem for derivatives.
• As a separate expression, there exists a [$ 0 < \theta < 1
• $f(a+h)=f(a)+h{f}^{\prime }(a+\theta h)$
  • Only the following rewrite is done $b=a+h,c=a+\theta h$.

By Lagrange’s mean value theorem $\log (1+x)=\frac{x}{1+\theta x}$

• because
• $a=1,h=x$ as
• $\log (1+x)=\log (1)+x{\log }^{\prime }(1+\theta x)$
• ${\log }^{\prime }(x)=1/x,\log (1)=0$, so
• $\log (1+x)=\frac{x}{1+\theta x}$
• Since log(x) is not differentiable at x=0, the absolute value of x must be less than 1

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