Condition of Constancy of the Function

Fact. Suppose function y=f(x){y}={f{{\left({x}\right)}}} is defined and continuous on interval X{X} and has finite derivative f(x){f{'}}{\left({x}\right)}. Function y=f(x){y}={f{{\left({x}\right)}}} is constant if and only if f(x){f{'}}{\left({x}\right)} for all x{x} in X{X}.

This fact means that if on some interval derivative of function equals 0 then function is constant their, its graph is just horizontal line.

Corollary. Suppose two function y=f(x){y}={f{{\left({x}\right)}}} and y=g(x){y}={g{{\left({x}\right)}}} are defined and continuous on interval X{X}, and have finite derivatives f(x){f{'}}{\left({x}\right)} and g(x){g{{\left({x}\right)}}}. Function y=f(x){y}={f{{\left({x}\right)}}} is constant if and only if f(x){f{'}}{\left({x}\right)} for all x{x} in X{X}. If f(x)=g(x){f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)} on interval X{X} then f(x)=g(x)+C{f{{\left({x}\right)}}}={g{{\left({x}\right)}}}+{C}, where C{C} is a constant for all x{x} in X{X}.

This corollary means that if functions have same derivatives on interval X{X} then their difference is constant.

Example 1. Consider functions f(x)=arctan(x){f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}} and g(x)=arcsin(x1+x2){g{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}.

They are defined and continuous on interval (,){\left(-\infty,\infty\right)}.

Since f(x)=11+x2{f{'}}{\left({x}\right)}=\frac{{1}}{{{1}+{{x}}^{{2}}}} and g(x)=11(x1+x2)21+x2x21+x21+x2=11+x2{g{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}^{{2}}}}\cdot\frac{{\sqrt{{{1}+{{x}}^{{2}}}}-\frac{{{{x}}^{{2}}}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}}}{{{1}+{{x}}^{{2}}}}=\frac{{1}}{{{1}+{{x}}^{{2}}}} then according to corollary arctan(x)=arcsin(x1+x2)+C{\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}+{C} on (,){\left(-\infty,\infty\right)}.

To find constant plug any value of x{x}, for example, x=0{x}={0}: arctan(0)=arcsin(01+02)+C{\operatorname{arctan}{{\left({0}\right)}}}={\operatorname{arcsin}{{\left(\frac{{0}}{\sqrt{{{1}+{{0}}^{{2}}}}}\right)}}}+{C} or C=0{C}={0}.

So, we have the following fact: arctan(x)=arcsin(x1+x2){\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}} for all x{x}.

Example 2. Consider functions f(x)=arctan(x){f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}} and g(x)=12arctan(2x1x2){g{{\left({x}\right)}}}=\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}.

It can be easily proven that f(x)=g(x){f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}. However, function g(x){g{{\left({x}\right)}}} is not defined when x=±1.{x}=\pm{1}. So, 12arctan(2x1x2)=arctan(x)+C\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+{C} on (,1),(1,1),(1,){\left(-\infty,-{1}\right)},{\left(-{1},{1}\right)},{\left({1},\infty\right)}.

It is interesting that constant will be different for different intervals.constancy of function

For interval (1,1){\left(-{1},{1}\right)} we plug x=0{x}={0}: 12arctan(20102)=arctan(0)+C\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}\cdot{0}}}{{{1}-{{0}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({0}\right)}}}+{C} or C=0{C}={0}.

For interval (,1){\left(-\infty,-{1}\right)} we let x{x}\to-\infty: 0=π2+C{0}=-\frac{\pi}{{2}}+{C} or C=π2{C}=\frac{\pi}{{2}}.

For interval (1,){\left({1},\infty\right)} we let x{x}\to\infty: 0=π2+C{0}=\frac{\pi}{{2}}+{C} or C=π2{C}=-\frac{\pi}{{2}}.

So,

12arctan(2x1x2)=arctan(x)+π2\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+\frac{\pi}{{2}} on (,1){\left(-\infty,-{1}\right)},

12arctan(2x1x2)=arctan(x)\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}} on (1,1){\left(-{1},{1}\right)},

12arctan(2x1x2)=arctan(x)π2\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}} on (1,){\left({1},\infty\right)}.

Graph confirms these facts.