Positive derivative implies increasing: Difference between revisions

Revision as of 20:42, 20 October 2011

Statement

On an open interval

Suppose $f$ is a function on an open interval $I$ that may be infinite in one or both directions (i..e, $I$ is of the form $\!(a,b)$ , $(a,\infty )$ , $(-\infty ,b)$ , or $(-\infty ,\infty )$ ). Suppose the derivative of $f$ exists and is positive everywhere on $I$ , i.e., $f'(x)>0$ for all $x\in I$ . Then, $f$ is an increasing function on $I$ , i.e.:

$\!x_{1},x_{2}\in I,\qquad x_{1}<x_{2}\implies f(x_{1})<f(x_{2})$

Facts used

Lagrange mean value theorem

Proof

Fill this in later

@@ Line 3: / Line 3: @@
 ===On an open interval===
-Suppose <math>f</math> is a function on an open interval <math>I</math> that may be infinite in one or both directions (i..e, <math>I</math> is of the form <math>(a,b)</math>, <math>(a,\infty)</math>, <math>(-\infty,b)</math>, or <math>(-\infty,\infty)</math>). Suppose the [[derivative]] of <math>f</math> exists and is positive everywhere on <math>I</math>, i.e., <math>f'(x) > 0</math> for all <math>x \in I</math>. Then, <math>f</math> is an [[fact about::increasing function]] on <math>I</math>, i.e.:
+Suppose <math>f</math> is a function on an open interval <math>I</math> that may be infinite in one or both directions (i..e, <math>I</math> is of the form <math>\! (a,b)</math>, <math>(a,\infty)</math>, <math>(-\infty,b)</math>, or <math>(-\infty,\infty)</math>). Suppose the [[derivative]] of <math>f</math> exists and is positive everywhere on <math>I</math>, i.e., <math>f'(x) > 0</math> for all <math>x \in I</math>. Then, <math>f</math> is an [[fact about::increasing function]] on <math>I</math>, i.e.:
-<math>x_1, x_2 \in I, \qquad x_1 < x_2 \implies f(x_1) < f(x_2)</math>
+<math>\! x_1, x_2 \in I, \qquad x_1 < x_2 \implies f(x_1) < f(x_2)</math>
 ==Facts used==