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Precalculus

3.8 Inverses and Radical Functions

Precalculus3.8 Inverses and Radical Functions
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  1. Preface
  2. 1 Functions
    1. Introduction to Functions
    2. 1.1 Functions and Function Notation
    3. 1.2 Domain and Range
    4. 1.3 Rates of Change and Behavior of Graphs
    5. 1.4 Composition of Functions
    6. 1.5 Transformation of Functions
    7. 1.6 Absolute Value Functions
    8. 1.7 Inverse Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  3. 2 Linear Functions
    1. Introduction to Linear Functions
    2. 2.1 Linear Functions
    3. 2.2 Graphs of Linear Functions
    4. 2.3 Modeling with Linear Functions
    5. 2.4 Fitting Linear Models to Data
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  4. 3 Polynomial and Rational Functions
    1. Introduction to Polynomial and Rational Functions
    2. 3.1 Complex Numbers
    3. 3.2 Quadratic Functions
    4. 3.3 Power Functions and Polynomial Functions
    5. 3.4 Graphs of Polynomial Functions
    6. 3.5 Dividing Polynomials
    7. 3.6 Zeros of Polynomial Functions
    8. 3.7 Rational Functions
    9. 3.8 Inverses and Radical Functions
    10. 3.9 Modeling Using Variation
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Review Exercises
    15. Practice Test
  5. 4 Exponential and Logarithmic Functions
    1. Introduction to Exponential and Logarithmic Functions
    2. 4.1 Exponential Functions
    3. 4.2 Graphs of Exponential Functions
    4. 4.3 Logarithmic Functions
    5. 4.4 Graphs of Logarithmic Functions
    6. 4.5 Logarithmic Properties
    7. 4.6 Exponential and Logarithmic Equations
    8. 4.7 Exponential and Logarithmic Models
    9. 4.8 Fitting Exponential Models to Data
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  6. 5 Trigonometric Functions
    1. Introduction to Trigonometric Functions
    2. 5.1 Angles
    3. 5.2 Unit Circle: Sine and Cosine Functions
    4. 5.3 The Other Trigonometric Functions
    5. 5.4 Right Triangle Trigonometry
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  7. 6 Periodic Functions
    1. Introduction to Periodic Functions
    2. 6.1 Graphs of the Sine and Cosine Functions
    3. 6.2 Graphs of the Other Trigonometric Functions
    4. 6.3 Inverse Trigonometric Functions
    5. Key Terms
    6. Key Equations
    7. Key Concepts
    8. Review Exercises
    9. Practice Test
  8. 7 Trigonometric Identities and Equations
    1. Introduction to Trigonometric Identities and Equations
    2. 7.1 Solving Trigonometric Equations with Identities
    3. 7.2 Sum and Difference Identities
    4. 7.3 Double-Angle, Half-Angle, and Reduction Formulas
    5. 7.4 Sum-to-Product and Product-to-Sum Formulas
    6. 7.5 Solving Trigonometric Equations
    7. 7.6 Modeling with Trigonometric Equations
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Review Exercises
    12. Practice Test
  9. 8 Further Applications of Trigonometry
    1. Introduction to Further Applications of Trigonometry
    2. 8.1 Non-right Triangles: Law of Sines
    3. 8.2 Non-right Triangles: Law of Cosines
    4. 8.3 Polar Coordinates
    5. 8.4 Polar Coordinates: Graphs
    6. 8.5 Polar Form of Complex Numbers
    7. 8.6 Parametric Equations
    8. 8.7 Parametric Equations: Graphs
    9. 8.8 Vectors
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  10. 9 Systems of Equations and Inequalities
    1. Introduction to Systems of Equations and Inequalities
    2. 9.1 Systems of Linear Equations: Two Variables
    3. 9.2 Systems of Linear Equations: Three Variables
    4. 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables
    5. 9.4 Partial Fractions
    6. 9.5 Matrices and Matrix Operations
    7. 9.6 Solving Systems with Gaussian Elimination
    8. 9.7 Solving Systems with Inverses
    9. 9.8 Solving Systems with Cramer's Rule
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  11. 10 Analytic Geometry
    1. Introduction to Analytic Geometry
    2. 10.1 The Ellipse
    3. 10.2 The Hyperbola
    4. 10.3 The Parabola
    5. 10.4 Rotation of Axes
    6. 10.5 Conic Sections in Polar Coordinates
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Review Exercises
    11. Practice Test
  12. 11 Sequences, Probability and Counting Theory
    1. Introduction to Sequences, Probability and Counting Theory
    2. 11.1 Sequences and Their Notations
    3. 11.2 Arithmetic Sequences
    4. 11.3 Geometric Sequences
    5. 11.4 Series and Their Notations
    6. 11.5 Counting Principles
    7. 11.6 Binomial Theorem
    8. 11.7 Probability
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  13. 12 Introduction to Calculus
    1. Introduction to Calculus
    2. 12.1 Finding Limits: Numerical and Graphical Approaches
    3. 12.2 Finding Limits: Properties of Limits
    4. 12.3 Continuity
    5. 12.4 Derivatives
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  14. A | Basic Functions and Identities
  15. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
  16. Index

Learning Objectives

In this section, you will:
  • Find the inverse of a polynomial function.
  • Restrict the domain to find the inverse of a polynomial function.

A mound of gravel is in the shape of a cone with the height equal to twice the radius.

Gravel in the shape of a cone.
Figure 1

The volume is found using a formula from elementary geometry.

V= 1 3 π r 2 h    = 1 3 π r 2 (2r)    = 2 3 π r 3 V= 1 3 π r 2 h    = 1 3 π r 2 (2r)    = 2 3 π r 3

We have written the volume V V in terms of the radius r. r. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula

r= 3V 2π 3 r= 3V 2π 3

This function is the inverse of the formula for V V in terms of r. r.

In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.

Finding the Inverse of a Polynomial Function

Two functions f f and g g are inverse functions if for every coordinate pair in f, (a, b), f, (a, b), there exists a corresponding coordinate pair in the inverse function, g,(b,a). g,(b,a). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.

For a function to have an inverse function the function to create a new function that is one-to-one and would have an inverse function.

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure 2. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Diagram of a parabolic trough that is 18” in height, 3’ in length, and 12” in width.
Figure 2

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x x measured horizontally and y y measured vertically, with the origin at the vertex of the parabola. See Figure 3.

Graph of a parabola.
Figure 3

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form y(x)=a x 2 . y(x)=a x 2 . Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a. a.

18=a 6 2  a= 18 36      = 1 2 18=a 6 2  a= 18 36      = 1 2

Our parabolic cross section has the equation

y(x)= 1 2 x 2 y(x)= 1 2 x 2

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y y the width will be given by 2x, 2x, so we need to solve the equation above for x x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x x values. On this domain, we can find an inverse by solving for the input variable:

y= 1 2 x 2 2y= x 2   x=± 2y y= 1 2 x 2 2y= x 2   x=± 2y

This is not a function as written. We are limiting ourselves to positive x x values, so we eliminate the negative solution, giving us the inverse function we’re looking for.

y= x 2 2 , x>0 y= x 2 2 , x>0

Because x x is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x. 2x. The trough is 3 feet (36 inches) long, so the surface area will then be:

Area=lw         =362x         =72x         =72 2y Area=lw         =362x         =72x         =72 2y

This example illustrates two important points:

  1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
  2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.

Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation f 1 (x). f 1 (x).

Warning: f 1 (x) f 1 (x) is not the same as the reciprocal of the function f( x ). f( x ). This use of “–1” is reserved to denote inverse functions. To denote the reciprocal of a function f( x ), f( x ), we would need to write ( f( x ) ) 1 = 1 f( x ) . ( f( x ) ) 1 = 1 f( x ) .

An important relationship between inverse functions is that they “undo” each other. If f 1 f 1 is the inverse of a function f, f, then f f is the inverse of the function f 1 . f 1 . In other words, whatever the function f f does to x, x, f 1 f 1 undoes it—and vice-versa. More formally, we write

f 1 ( f( x ) )=x,for all x in the domain of f f 1 ( f( x ) )=x,for all x in the domain of f

and

f( f 1 ( x ) )=x,for all x in the domain of  f 1 f( f 1 ( x ) )=x,for all x in the domain of  f 1

Verifying Two Functions Are Inverses of One Another

Two functions, f f and g, g, are inverses of one another if for all x x in the domain of f f and g. g.

g( f( x ) )=f( g( x ) )=x g( f( x ) )=f( g( x ) )=x

How To

Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.

  1. Replace f( x ) f( x ) with y. y.
  2. Interchange x x and y. y.
  3. Solve for y, y, and rename the function f 1 (x). f 1 (x).

Example 1

Verifying Inverse Functions

Show that f( x )= 1 x+1 f( x )= 1 x+1 and f 1 ( x )= 1 x 1 f 1 ( x )= 1 x 1 are inverses, for x0,1 x0,1 .

Try It #1

Show that f( x )= x+5 3 f( x )= x+5 3 and f 1 ( x )=3x5 f 1 ( x )=3x5 are inverses.

Example 2

Finding the Inverse of a Cubic Function

Find the inverse of the function f(x)=5 x 3 +1. f(x)=5 x 3 +1.

Analysis

Look at the graph of f f and f 1 . f 1 . Notice that the two graphs are symmetrical about the line y=x. y=x. This is always the case when graphing a function and its inverse function.

Also, since the method involved interchanging x x and y, y, notice corresponding points. If (a,b) (a,b) is on the graph of f, f, then (b,a) (b,a) is on the graph of f 1 . f 1 . Since (0,1) (0,1) is on the graph of f, f, then (1,0) (1,0) is on the graph of f 1 . f 1 . Similarly, since (1,6) (1,6) is on the graph of f, f, then (6,1) (6,1) is on the graph of f 1 . f 1 . See Figure 4.

Graph of f(x)=5x^3+1 and its inverse, f^(-1)(x)=3sqrt((x-1)/(5)).
Figure 4
Try It #2

Find the inverse function of f(x)= x+4 3 . f(x)= x+4 3 .

Restricting the Domain to Find the Inverse of a Polynomial Function

So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.

Restricting the Domain

If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.

How To

Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

  1. Restrict the domain by determining a domain on which the original function is one-to-one.
  2. Replace f(x) with y. f(x) with y.
  3. Interchange x and y. x and y.
  4. Solve for y, y, and rename the function or pair of function f 1 (x). f 1 (x).
  5. Revise the formula for f 1 (x) f 1 (x) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.

Example 3

Restricting the Domain to Find the Inverse of a Polynomial Function

Find the inverse function of f: f:

  1. f(x)= (x4) 2 , x4 f(x)= (x4) 2 , x4
  2. f(x)= (x4) 2 , x4 f(x)= (x4) 2 , x4

Analysis

On the graphs in Figure 6, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line y=x. y=x. The coordinate pair (4, 0) (4, 0) is on the graph of f f and the coordinate pair (0, 4) (0, 4) is on the graph of f 1 . f 1 . For any coordinate pair, if ( a, b ) ( a, b ) is on the graph of f, f, then ( b, a ) ( b, a ) is on the graph of f 1 . f 1 . Finally, observe that the graph of f f intersects the graph of f 1 f 1 on the line y=x. y=x. Points of intersection for the graphs of f f and f 1 f 1 will always lie on the line y=x. y=x.

Two graphs of a parabolic function with half of its inverse.
Figure 6

Example 4

Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified

Restrict the domain and then find the inverse of

f(x)= (x2) 2 3. f(x)= (x2) 2 3.

Analysis

Notice that we arbitrarily decided to restrict the domain on x2. x2. We could just have easily opted to restrict the domain on x2, x2, in which case f 1 (x)=2 x+3 . f 1 (x)=2 x+3 . Observe the original function graphed on the same set of axes as its inverse function in Figure 7. Notice that both graphs show symmetry about the line y=x. y=x. The coordinate pair ( 2, 3 ) ( 2, 3 ) is on the graph of f f and the coordinate pair ( 3, 2 ) ( 3, 2 ) is on the graph of f 1 . f 1 . Observe from the graph of both functions on the same set of axes that

domain of f=range of  f 1 =[ 2, ) domain of f=range of  f 1 =[ 2, )

and

domain of  f 1 =range of f=[ 3, ) domain of  f 1 =range of f=[ 3, )

Finally, observe that the graph of f f intersects the graph of f 1 f 1 along the line y=x. y=x.

Graph of a parabolic function with half of its inverse.
Figure 7
Try It #3

Find the inverse of the function f(x)= x 2 +1, f(x)= x 2 +1, on the domain x0. x0.

Solving Applications of Radical Functions

Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited.

How To

Given a radical function, find the inverse.

  1. Determine the range of the original function.
  2. Replace f( x ) f( x ) with y, y, then solve for x. x.
  3. If necessary, restrict the domain of the inverse function to the range of the original function.

Example 5

Finding the Inverse of a Radical Function

Restrict the domain and then find the inverse of the function f(x)= x4 . f(x)= x4 .

Analysis

Notice in Figure 8 that the inverse is a reflection of the original function over the line y=x. y=x. Because the original function has only positive outputs, the inverse function has only positive inputs.

Graph of f(x)=sqrt(x-4) and its inverse, f^(-1)(x)=x^2+4.
Figure 8

Try It #4

Restrict the domain and then find the inverse of the function f(x)= 2x+3 . f(x)= 2x+3 .

Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.

Example 6

Solving an Application with a Cubic Function

A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

V= 2 3 π r 3 V= 2 3 π r 3

Find the inverse of the function V= 2 3 π r 3 V= 2 3 π r 3 that determines the volume V V of a cone and is a function of the radius r. r. Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use π=3.14. π=3.14.

Determining the Domain of a Radical Function Composed with Other Functions

When radical functions are composed with other functions, determining domain can become more complicated.

Example 7

Finding the Domain of a Radical Function Composed with a Rational Function

Find the domain of the function f(x)= (x+2)(x3) (x1) . f(x)= (x+2)(x3) (x1) .

Finding Inverses of Rational Functions

As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.

Example 8

Finding the Inverse of a Rational Function

The function C= 20+0.4n 100+n C= 20+0.4n 100+n represents the concentration C C of an acid solution after n n mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for n n in terms of C. C. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.

Try It #5

Find the inverse of the function f(x)= x+3 x2 . f(x)= x+3 x2 .

3.8 Section Exercises

Verbal

1.

Explain why we cannot find inverse functions for all polynomial functions.

2.

Why must we restrict the domain of a quadratic function when finding its inverse?

3.

When finding the inverse of a radical function, what restriction will we need to make?

4.

The inverse of a quadratic function will always take what form?

Algebraic

For the following exercises, find the inverse of the function on the given domain.

5.

f( x )= ( x4 ) 2 , [4,) f( x )= ( x4 ) 2 , [4,)

6.

f( x )= ( x+2 ) 2 , [2,) f( x )= ( x+2 ) 2 , [2,)

7.

f(x)= ( x+1 ) 2 3, [1,) f(x)= ( x+1 ) 2 3, [1,)

8.

f(x)=2 3+x f(x)=2 3+x

9.

f(x)=3 x 2 +5,( ,0 ] f(x)=3 x 2 +5,( ,0 ]

10.

f( x )=12 x 2 , [0,) f( x )=12 x 2 , [0,)

11.

f( x )=9 x 2 , [0,) f( x )=9 x 2 , [0,)

12.

f(x)=2 x 2 +4, [0,) f(x)=2 x 2 +4, [0,)

For the following exercises, find the inverse of the functions.

13.

f(x)= x 3 +5 f(x)= x 3 +5

14.

f( x )=3 x 3 +1 f( x )=3 x 3 +1

15.

f(x)=4 x 3 f(x)=4 x 3

16.

f( x )=42 x 3 f( x )=42 x 3

For the following exercises, find the inverse of the functions.

17.

f(x)= 2x+1 f(x)= 2x+1

18.

f(x)= 34x f(x)= 34x

19.

f( x )=9+ 4x4 f( x )=9+ 4x4

20.

f( x )= 6x8 +5 f( x )= 6x8 +5

21.

f( x )=9+2 x 3 f( x )=9+2 x 3

22.

f( x )=3 x 3 f( x )=3 x 3

23.

f( x )= 2 x+8 f( x )= 2 x+8

24.

f( x )= 3 x4 f( x )= 3 x4

25.

f( x )= x+3 x+7 f( x )= x+3 x+7

26.

f( x )= x2 x+7 f( x )= x2 x+7

27.

f( x )= 3x+4 54x f( x )= 3x+4 54x

28.

f( x )= 5x+1 25x f( x )= 5x+1 25x

29.

f(x)= x 2 +2x, [1,) f(x)= x 2 +2x, [1,)

30.

f(x)= x 2 +4x+1, [2,) f(x)= x 2 +4x+1, [2,)

31.

f(x)= x 2 6x+3, [3,) f(x)= x 2 6x+3, [3,)

Graphical

For the following exercises, find the inverse of the function and graph both the function and its inverse.

32.

f(x)= x 2 +2,x0 f(x)= x 2 +2,x0

33.

f(x)=4 x 2 ,x0 f(x)=4 x 2 ,x0

34.

f(x)= ( x+3 ) 2 ,x3 f(x)= ( x+3 ) 2 ,x3

35.

f(x)= ( x4 ) 2 ,x4 f(x)= ( x4 ) 2 ,x4

36.

f(x)= x 3 +3 f(x)= x 3 +3

37.

f(x)=1 x 3 f(x)=1 x 3

38.

f(x)= x 2 +4x,x2 f(x)= x 2 +4x,x2

39.

f(x)= x 2 6x+1,x3 f(x)= x 2 6x+1,x3

40.

f(x)= 2 x f(x)= 2 x

41.

f(x)= 1 x 2 ,x0 f(x)= 1 x 2 ,x0

For the following exercises, use a graph to help determine the domain of the functions.

42.

f(x)= (x+1)(x1) x f(x)= (x+1)(x1) x

43.

f(x)= (x+2)(x3) x1 f(x)= (x+2)(x3) x1

44.

f(x)= x(x+3) x4 f(x)= x(x+3) x4

45.

f(x)= x 2 x20 x2 f(x)= x 2 x20 x2

46.

f(x)= 9 x 2 x+4 f(x)= 9 x 2 x+4

Technology

For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with y-coordinates given.

47.

f(x)= x 3 x2,y=1, 2, 3 f(x)= x 3 x2,y=1, 2, 3

48.

f(x)= x 3 +x2, y=0, 1, 2 f(x)= x 3 +x2, y=0, 1, 2

49.

f(x)= x 3 +3x4, y=0, 1, 2 f(x)= x 3 +3x4, y=0, 1, 2

50.

f(x)= x 3 +8x4, y=1, 0, 1 f(x)= x 3 +8x4, y=1, 0, 1

51.

f(x)= x 4 +5x+1, y=1, 0, 1 f(x)= x 4 +5x+1, y=1, 0, 1

Extensions

For the following exercises, find the inverse of the functions with a,b,c a,b,c positive real numbers.

52.

f(x)=a x 3 +b f(x)=a x 3 +b

53.

f(x)= x 2 +bx f(x)= x 2 +bx

54.

f(x)= a x 2 +b f(x)= a x 2 +b

55.

f(x)= ax+b 3 f(x)= ax+b 3

56.

f(x)= ax+b x+c f(x)= ax+b x+c

Real-World Applications

For the following exercises, determine the function described and then use it to answer the question.

57.

An object dropped from a height of 200 meters has a height, h( t ), h( t ), in meters after t t seconds have lapsed, such that h(t)=2004.9 t 2 . h(t)=2004.9 t 2 . Express t t as a function of height, h, h, and find the time to reach a height of 50 meters.

58.

An object dropped from a height of 600 feet has a height, h( t ), h( t ), in feet after t t seconds have elapsed, such that h(t)=60016 t 2 . h(t)=60016 t 2 . Express t t as a function of height h, h, and find the time to reach a height of 400 feet.

59.

The volume, V, V, of a sphere in terms of its radius, r, r, is given by V(r)= 4 3 π r 3 . V(r)= 4 3 π r 3 . Express r r as a function of V, V, and find the radius of a sphere with volume of 200 cubic feet.

60.

The surface area, A, A, of a sphere in terms of its radius, r, r, is given by A(r)=4π r 2 . A(r)=4π r 2 . Express r r as a function of V, V, and find the radius of a sphere with a surface area of 1000 square inches.

61.

A container holds 100 ml of a solution that is 25 ml acid. If n n ml of a solution that is 60% acid is added, the function C(n)= 25+.6n 100+n C(n)= 25+.6n 100+n gives the concentration, C, C, as a function of the number of ml added, n. n. Express n n as a function of C C and determine the number of mL that need to be added to have a solution that is 50% acid.

62.

The period T, T, in seconds, of a simple pendulum as a function of its length l, l, in feet, is given by T(l)=2π l 32.2 T(l)=2π l 32.2 . Express l l as a function of T T and determine the length of a pendulum with period of 2 seconds.

63.

The volume of a cylinder , V, V, in terms of radius, r, r, and height, h, h, is given by V=π r 2 h. V=π r 2 h. If a cylinder has a height of 6 meters, express the radius as a function of V V and find the radius of a cylinder with volume of 300 cubic meters.

64.

The surface area, A, A, of a cylinder in terms of its radius, r, r, and height, h, h, is given by A=2π r 2 +2πrh. A=2π r 2 +2πrh. If the height of the cylinder is 4 feet, express the radius as a function of V V and find the radius if the surface area is 200 square feet.

65.

The volume of a right circular cone, V, V, in terms of its radius, r, r, and its height, h, h, is given by V= 1 3 π r 2 h. V= 1 3 π r 2 h. Express r r in terms of V V if the height of the cone is 12 inches and find the radius of a cone with volume of 50 cubic inches.

66.

Consider a cone with height of 30 feet. Express the radius, r, r, in terms of the volume, V, V, and find the radius of a cone with volume of 1000 cubic feet.

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