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Let \(f\) be a function defined on \([a, b]\) such that \(f' (x) > 0\), for all \(x \: \in (a, b)\). Then prove that \(f\) is an increasing function on \((a, b).\)

1 Answer

  • $f'(c)=\large\frac{f(x_2)-f(x_1)}{x_2-x_1}$
Step 1:
Let $x_1,x_2\in(a,b)$ such that $x_1 < x_2$
Consider the sub-interval $[x_1,x_2]$
Since $f(x)$ is differentiate on $(a,b)$ and $[x_1,x_2]\in (a,b)$
Therefore $f(x)$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$
By the Lagrange's mean value theorem ,there exists $c\in (x_1,x_2)$
Such that $f'(c)=\large\frac{f(x_2)-f(x_1)}{x_2-x_1}$
Step 2:
Since $f'(x) >0$ for all $x\in (a,b)$.
So in particular $f'(c)>0$
Now $f'(c) >0$
$\Rightarrow \large\frac{f(x_2)-f(x_1)}{x_2-x_1} $$>0$
$\Rightarrow f(x_2)-f(x_1)>0$
$x_2-x_1>0$ when $x_1 < x_2$
$f(x_1) < f(x_2)$ if $x_1$ < $x_2$
Since $x_1,x_2$ are arbitrary points in $(a,b)$
$\therefore x_1 < x_2$
$\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2 \in (a,b)$
Hence $f(x)$ is increasing on $(a,b)$
answered Aug 12, 2013 by sreemathi.v

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