Positive derivative implies increasing: Difference between revisions

Latest revision as of 02:11, 25 September 2021

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

Opposite facts

Positive derivative at a point not implies increasing around the point

Facts used

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] Since $f$ is defined and continuous on an interval containing both $x_{1}$ and $x_{2}$ , it is in particular defined and on $[x_{1},x_{2}]$ , which lies inside the open interval. Further, $f$ is differentiable on the interior of $I$ , and hence on the open interval $(x_{1},x_{2})$ , which is contained in the interior of $I$ . Thus, the conditions needed to apply Fact (1) are available. Using Fact (1) gives the existence of $x_{3}$ such that $f'(x_{3})$ equals the difference quotient.
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] By Step (1), there exists $x_{3}$ such that $f'(x_{3})$ equals the difference quotient. From the given data, $f'(x_{3})$ is positive. Combining, we obtain that the difference quotient itself is positive.
3	$f(x_{1})<f(x_{2})$		$x_{1}<x_{2}$	Step (2)	[SHOW MORE] In Step (2), we obtained that the difference quotient is positive. The denominator of the expression is positive because $x_{1}<x_{2}$ . Thus, the numerator must also be positive, giving $f(x_{1})<f(x_{2})$ upon rearrangement.

@@ 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, x_1 < x_2 \implies f(x_1) < f(x_2)</math>
+===On a general interval===
+Suppose <math>f</math> is a function on an interval <math>I</math> that may be infinite in one or both directions and may be open or closed at either end. Suppose <math>f</math> is a [[continuous function]] on all of <math>I</math> and that the [[derivative]] of <math>f</math> exists and is positive everywhere on the interior of <math>I</math>, i.e., <math>f'(x) > 0</math> for all <math>x \in I</math> other than the endpoints of <math>I</math> (if they exist). 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>
+==Related facts==
+===Similar facts===
+* [[Zero derivative implies locally constant]]
+* [[Negative derivative implies decreasing]]
+* [[Nonnegative derivative that is zero only at isolated points implies increasing]]
+* [[Increasing and differentiable implies nonnegative derivative]]
+===Opposite facts===
+* [[Positive derivative at a point not implies increasing around the point]]
 ==Facts used==
@@ Line 13: / Line 32: @@
 ==Proof==
-{{fillin}}
+===General version===
+'''Given''': A function <math>f</math> on interval <math>I</math> such that <math>f'(x) > 0</math> for all <math>x</math> in the interior of <math>I</math> and <math>f</math> is continuous on <math>I</math>. Numbers <math>x_1 < x_2</math> with <math>x_1, x_2 \in I</math>.
+'''To prove''': <math>\! f(x_1) < f(x_2)</math>
+'''Proof''':
+{| class="sortable" border="1"
+! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
+|-
+| 1 || Consider the [[difference quotient]] <math>\! \frac{f(x_2) - f(x_1)}{x_2 - x_1}</math>. There exists <math>x_3</math> such that <math>x_1 < x_3 < x_2</math> and <math>\! f'(x_3)</math> equals this difference quotient. || Fact (1) || <math>x_1 < x_2</math>, <math>f</math> is defined and continuous on an interval <math>I</math> containing <math>x_1,x_2</math>, differentiable on the interior of the interval. || || <toggledisplay>Since <math>f</math> is defined and continuous on an interval containing both <math>x_1</math> and <math>x_2</math>, it is in particular defined and on <math>[x_1,x_2]</math>, which lies inside the open interval. Further, <math>f</math> is differentiable on the interior of <math>I</math>, and hence on the open interval <math>(x_1,x_2)</math>, which is contained in the interior of <math>I</math>. Thus, the conditions needed to apply Fact (1) are available. Using Fact (1) gives the existence of <math>x_3</math> such that <math>f'(x_3)</math> equals the difference quotient.</toggledisplay>
+|-
+| 2 || The difference quotient <math>\! \frac{f(x_2) - f(x_1)}{x_2 - x_1}</math> is positive. || ||<math>f'(x)</math> is positive for all <math>x</math> in the interior of <math>I</math>. || Step (1) || <toggledisplay>By Step (1), there exists <math>x_3</math> such that <math>f'(x_3)</math> equals the difference quotient. From the given data, <math>f'(x_3)</math> is positive. Combining, we obtain that the difference quotient itself is positive.</toggledisplay>
+|-
+| 3 || <math>f(x_1) < f(x_2)</math> || || <math>x_1 < x_2</math> || Step (2) || <toggledisplay>In Step (2), we obtained that the difference quotient is positive. The denominator of the expression is positive because <math>x_1 < x_2</math>. Thus, the numerator must also be positive, giving <math>f(x_1) < f(x_2)</math> upon rearrangement.</toggledisplay>
+|}