Difference between revisions of "Positive derivative implies increasing"

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===On an open interval===
 
===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, x_1 < x_2 \implies f(x_1) < f(x_2)</math>
 
<math>x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)</math>

Latest revision as of 16:56, 13 December 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, x_1 < x_2 \implies f(x_1) < f(x_2)

On a general interval

Suppose f is a function on an interval I that may be infinite in one or both directions and may be open or closed at either end. Suppose f is a continuous function on all of I and that the derivative of f exists and is positive everywhere on the interior of I, i.e., f'(x) > 0 for all x \in I other than the endpoints of I (if they exist). Then, f is an increasing function on I, i.e.:

x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)

Related facts

Similar facts

Facts used

  1. Lagrange mean value theorem

Proof

General version

Given: A function f on interval I such that f'(x) > 0 for all x in the interior of I and f is continuous on I. Numbers x_1 < x_2 with x_1, x_2 \in I.

To prove: \! f(x_1) < f(x_2)

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the difference quotient \! \frac{f(x_2) - f(x_1)}{x_2 - x_1}. There exists x_3 such that x_1 < x_3 < x_2 and \! f'(x_3) equals this difference quotient. Fact (1) x_1 < x_2, f is defined and continuous on an interval I containing x_1,x_2, differentiable on the interior of the interval. [SHOW MORE]
2 The difference quotient \! \frac{f(x_2) - f(x_1)}{x_2 - x_1} is positive. f'(x) is positive for all x in the interior of I. Step (1) [SHOW MORE]
3 f(x_1) < f(x_2) x_1 < x_2 Step (2) [SHOW MORE]