Graph, Domain và Range of arccos(x) function

The graph, domain và range and other properties of the inverse trigonometric function ( arccos(x) ) are explored using graphs, examples with detailed solutions & an interactive app.

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Set the parameters to lớn ( a = 1, b = 1, c = 0) và ( d = 0 ) lớn obtain< f(x) = arccos(x) >Check that the tên miền of ( arccos(x) ) is given by the interval ( <-1 , 1> ) & the range is given by the interval ( <0 , pi > ) , ( (pi approx 3.14) ).Change coefficient ( a ) và explore its affect on the range of ( a arccos(x) ) function. (Hint: vertical compression, stretching, reflection).Does the tên miền of ( a arccos(x) ) change with ( a ).?Does a change in ( b ) affect the domain or/and the range of ( a arccos(b x) )? (Hint: horizontal compression, stretching).Does a change in ( c ) affect the tên miền or/and the range of ( aarccos(b x + c) )? How about horizontal shift?Does a change in ( d ) affect the range of ( aarccos(bx + c) + d )?What is the domain và range of ( a arccos(b x + c) + d ) in terms of ( a, b, c ) and ( d )?

## Exercises

Find the domain và range of ( f(x) = arccos(2 x + 5) - pi/2 ).Find the domain and range of ( g(x) = - 0.5 arccos(x + 4) - pi ).Find the domain & range of ( h(x) = - dfrac12 arccos(x) - pi / 4).Answers khổng lồ Above QuestionsDomain: ( <-3 , -2> ) , Range: ( <- pi/2 , pi/2> ).Domain: ( <-5 , -3> ) , Range: ( < -3pi / 2 , -pi> ).Domain: ( <-1 , 1> ) , Range: ( <- 3pi/4 , pi/4> ).

## More References and link to Inverse Trigonometric Functions

Find Domain & Range of Arccosine FunctionsInverse Trigonometric FunctionsGraph, Domain and Range of Arcsin functionGraph, Domain và Range of Arctan functionFind Domain và Range of Arcsine FunctionsSolve Inverse Trigonometric Functions Questions

## Definition of arccos(x) Functions

The function ( cos(x) ) is shown below. On its implied domain, cos(x) is not a one khổng lồ one functionas seen below; a horizontal line chạy thử for a one to one function would fail. But we limit the domain to ( < 0 , pi > ), blue graph below, we obtain a one to lớn one function that has an inverse which cannot be obtained algebraically.   Example 1 Evaluate ( arccos(x) ) given the value of ( x ).Special values related lớn special angles( arccos(1) = 0) because ( cos(0) = 1 )( arccos(-1) = pi ; ext or 180^o ) because ( cos(pi) = - 1 )( arccos(2) = ) undefined because ( 2 ) is not in the domain name of ( arccos(x) ) which is ( -1 le x le 1 ) ( there is no angle that has cosine equal lớn 2 ).( arccos(-dfrac12) = dfrac2pi3 ; ext or 120^o ) because ( cos(dfrac2pi3) = -dfrac12 )( arccos( - dfracsqrt32) = dfrac5pi6 ; ext or 150^o) because ( cos(dfrac5pi6) = - dfracsqrt32 )( arccos(dfracsqrt 32) = dfracpi6 ; ext or 30^o ) because ( sin(dfracpi6) = dfracsqrt 32 )Use of calculator( arccos(0.25) = 1.32 ; ext or 75.52^o )( arccos(-0.77) = 2.45 ; ext or 140.35^o )

## Properties of ( y = arccos(x) )

Domain: ( <-1 , 1> )Range: ( <0 , pi> )( arccos(x) ) is a one lớn one function
( cos(arccos(x)) = x ) , for x in the interval ( <-1,1> ) , due to lớn property of a function and its inverse :( f(f^-1(x) = x ) where ( x ) is in the tên miền of ( f^-1 )( arccos(cos(x)) = x ) , for x in the interval ( <0 , pi > ) , due to lớn property of a function và its inverse :( f^-1(f(x) = x ) where ( x ) is in the domain of ( f )Example 2 Find the domain name of the functions:a) ( y = arccos(-3x)) b) ( y = arccos(2 x - 1) ) c) ( y = 3 arccos(x/2 + 1) + pi/4 )Solution to lớn Example 2a)the domain name is found by first writing that the argument ( -3x ) of the given function ( y = arccos(-3x)) is within the tên miền of the arccos function which one of the properties given above. Hence the need to solve the double inequality( -1 le -3x le 1 )divide all terms of the double inequality by - 3 và change the symbols of inequality khổng lồ obtain( - 1/3 le x le 1/3 ) , which is the domain name of the function ( y = arccos(-3x)).b)Solve the inequality( -1 le 2 x - 1 le 1 )( 0 le 2 x le 2 )( 0 le x le 1 ) , which is the domain of the function ( y = arccos(2 x - 1) ).c)( -1 le x/2 + 1 le 1 )Solve the above inequality( - 4 le x le 0 ) , which is the domain of the function ( y = 3 arccos(x/2 + 1) + pi/4 ).Example 3 Find the range of the functions:a) ( y = - 2 arccos(x)) b) ( y = - arccos(x) + pi/4 ) c) ( y = - arccos(x-1) - pi)Solution to lớn Example 3a)We start with the range of ( arccos(x)) as a double inequality( 0 le arccos(x) le pi )multiply all terms of the above inequality by - 2, change the symbols of inequality and simplify( - 2 pi le - 2arccos(x) le 0 )the range of the given function ( 2 arccos(x) ) is given by the interval ( < - 2 pi , 0 > ).b)we start with the range of ( arccos(x))( 0 le arccos(x) le pi )Multiply all terms of the above inequality by -1 và change symbol of the double inequality( -pi le -arccos(x) le 0 )add ( dfracpi4 ) khổng lồ all terms of the above inequality & simplify( - 3 dfracpi4 le - arccos(x) + dfracpi4 le dfracpi4 )the range of the given function ( y = - arccos(x) + pi/4 ) is given by the interval ( < - 3 dfracpi4 , dfracpi4 > ).c)The graph of the given function ( arccos(x-1)) is the graph of ( arccos(x)) shifted 1 unit to the right. Shifting a graph lớn the left or to the right does not affect the range. Hencethe range of ( arccos(x-1)) is given by the interval ( < 0 , pi > ) and may be written as a double inequality( 0 le arccos(x-1) le pi )Mutliply all terms of the inequality by -1 & change the symbol of inequality( -pi le -arccos(x-1)le 0 )Add ( -pi ) lớn all terms of the inequality và simplify to lớn obtain the range of the function ( y = - arccos(x-1) - pi)( - 2 pi le - arccos(x-1) -pi le -pi )Example 4 Evaluate if possiblea) ( cos(arccos(-0.5))) b) ( arccos(cos(dfracpi6) ) ) c) ( cos(arccos(-2.1))) d) ( arccos(cos(dfrac- pi3) ) )Solution to Example 4a)( cos(arccos(-0.5)) = -0.5) using property 4 aboveb)( arccos(cos(dfracpi6) ) = dfracpi6) using property 5 abovec)NOTE that we cannot use property 4 because -2.1 is not in the tên miền of ( arccos(x) )( cos(arccos(-2.1)) is undefinedd)NOTE that we cannot use property 5 because ( - dfracpi3 ) is not in the domain name of that property. We will first transform ( cos(-dfracpi3) ) as follows( cos(- dfracpi3) = cos(dfracpi3)) , cosine function is evenWe now substitute ( cos( - dfrac pi3) ) by cos(dfracpi3) in the given expression( arccos(cos( - dfracpi3) ) = arccos(cos(dfracpi3) ))We now use the property 5 because ( dfracpi3 ) is in the domain of the property. Hence( arccos(cos( - dfracpi3) ) = arccos(cos(dfracpi3)) = dfracpi3 )

## Interactive Tutorial lớn Explore the Transformed arccos(x)

The exploration is carried out by analyzing the effects of the parameters ( a, b, c) và ( d ) included in the more general arccos function given by< f(x) = a arccos(b x + c) + d >Change parameters ( a, b, c) and ( d ) & click on the button "draw" in the left panel below. Zoom in & out for better viewing.

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