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.

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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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a = 1