﻿ ﻿Big Idea 2: Derivative. Differentiation - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Calculus AB 2017 (2016)

## 5 Steps to a 5: AP Calculus AB 2017 (2016)

### Differentiation

IN THIS CHAPTER

Summary: The derivative of a function is often used to find rates of change. It is also related to the slope of a tangent line. On the AP Calculus AB exam, many questions involve finding the derivative of a function. In this chapter, you will learn different techniques for finding a derivative which include using the Power Rule, Product and Quotient Rules, Chain Rule, and Implicit Differentiation. You will also learn to find the derivatives of trigonometric, exponential, logarithmic, and inverse functions.

Key Ideas

Definition of the derivative of a function

Power Rule, Product and Quotient Rules, and Chain Rule

Derivatives of trigonometric, exponential, and logarithmic functions

Derivatives of inverse functions

Implicit Differentiation

Higher order derivatives

L’Hôpital’s Rule for Indeterminate Forms

7.1 Derivatives of Algebraic Functions

Main Concepts: Definition of the Derivative of a Function; Power Rule; The Sum, Difference, Product, and Quotient Rules; The Chain Rule

Definition of the Derivative of a Function

The derivative of a function f , written as f ′, is defined as

if this limit exists. (Note that f ′(x ) is read as f prime of x .)

Other symbols of the derivative of a function are:

.

Let m tangent be the slope of the tangent to a curve y = f (x ) at a point on the curve. Then,

.

(See Figure 7.1-1 .)

Figure 7.1-1

Given a function f , if f ′(x ) exists at x = a , then the function f is said to be differentiable at x = a . If a function f is differentiable at x = a , then f is continuous at x = a . (Note that the converse of the statement is not necessarily true, i.e., if a function f is continuous at x = a , then f may or may not be differentiable at x = a .) Here are several examples of functions that are not differentiable at a given number x = a . (See Figures 7.1-27.1-5 .)

Figure 7.1-2

Figure 7.1-3

Figure 7.1-4

Figure 7.1-5

Example 1

If f (x ) = x 2 – 2x – 3, find (a) f ′(x ) using the definition of derivative, (b) f ′(0), (c) f ′(1), and (d) f ′(3).

(a) Using the definition of derivative,

(b) f ′(0) = 2(0) – 2 = –2, (c) f ′(1) = 2(1) – 2 = 0, and (d) f ′(3) = 2(3) – 2 = 4.

Example 2

Evaluate .

The expression is equivalent to the derivative of the function f (x ) = cos x at x = π , i.e., f ′(π ). The derivative of f (x ) = cos x at x = π is equivalent to the slope of the tangent to the curve of cos x at x= π . The tangent is parallel to the x -axis. Thus, the slope is 0 or = 0.

Or, using an algebraic method, note that cos(a + b ) = cos(a ) cos(b ) – sin(a ) sin(b ). Then rewrite .

(See Figure 7.1-6 .)

Figure 7.1-6

Example 3

If the function f (x ) = x 2/3 + 1, find all points where f is not differentiable.

The function f (x ) is continuous for all real numbers and the graph of f (x ) forms a “cusp” at the point (0, 1). Thus, f (x ) is not differentiable at x = 0. (See Figure 7.1-7 .)

Figure 7.1-7

Example 4

Using a calculator, find the derivative of f (x ) = x 2 + 4x at x = 3.

There are several ways to find f ′(3), using a calculator. One way is to use the [nDeriv ] function of the calculator. From the main Home screen, select F3-Calc and then select [nDeriv ]. Enter [nDeriv ] (x 2 + 4x , x )|x = 3. The result is 10.

• Always write out all formulas in your solutions.

Power Rule

If f (x ) = c where c is a constant, then f ′(x ) = 0.

If f (x ) = xn where n is a real number, then f ′(x ) = nx n –1 .

If f (x ) = c x n where c is a constant and n is a real number, then f ′(x ) = cnx n –1 .

Summary of Derivatives of Algebraic Functions

Example 1

If f (x ) = 2x 3 , find (a) f ′(x ), (b) f ′(1), and (c) f ′(0).

Note that (a) f ′(x ) = 6x 2 , (b) f ′(1) = 6(1)2 = 6, and (c) f ′(0) = 0.

Example 2

If , find (a) and (b) (which represents at x = 0).

Note that (a) and thus, and (b) does not exist because the expression is undefined.

Example 3

Here are several examples of algebraic functions and their derivatives:

Example 4

Using a calculator, find f ′(x ) and f ′(3) if .

There are several ways of finding f ′(x ) and f ′(9) using a calculator. One way is to use the d [Differentiate ] function. Go to the Home screen. Select F3-Calc and then select d [Differentiate ]. Enter . The result is . To find f ′(3), enter x )|x = 3. The result is .

The Sum, Difference, Product, and Quotient Rules

If u and v are two differentiable functions, then

Example 1

Find f ′(x ) if f (x ) = x 3 – 10x + 5.

Using the sum and difference rules, you can differentiate each term and obtain f ′(x ) = 3x 2 – 10. Or using your calculator, select the d [Differentiate ] function and enter d (x 3 – 10x + 5, x ) and obtain 3x 2 – 10.

Example 2

If y = (3x – 5)(x 4 + 8x – 1), find .

Using the product rule , let u = (3x – 5) and v = (x 4 + 8x – 1).

Then = (3)(x 4 + 8x – 1) + (4x 3 + 8)(3x – 5) = (3x 4 + 24x – 3) + (12x 4 – 20x 3 + 24x – 40) = 15x 4 – 20x 3 + 48x – 43. Or you can use your calculator and enter d ((3x – 5)(x 4 + 8x – 1), x ) and obtain the same result.

Example 3

If , find f ′(x ).

Using the quotient rule , let u = 2x – 1 and v = x + 5. Then . Or you can use your calculator and enter d ((2x – 1)/(x + 5), x ) and obtain the same result.

Example 4

Using your calculator, find an equation of the tangent to the curve f (x ) = x 2 – 3x + 2 at x = 5.

Find the slope of the tangent to the curve at x = 5 by entering d (x 2 – 3x + 2, x )|x = 5. The result is 7. Compute f (5) = 12. Thus, the point (5, 12) is on the curve of f (x ). An equation of the line whose slope m = 7 and passing through the point (5, 12) is y – 12 = 7(x – 5).

• Remember that ln and . The integral formula is not usually tested in the AB exam.

The Chain Rule

If y = f (u ) and u = g (x ) are differentiable functions of u and x respectively, then or .

Example 1

If y = (3x – 5)10 , find .

Using the chain rule, let u = 3x – 5 and thus, y = u 10 . Then, and .

Since (3) = 10(3x – 5)9 (3) = 30(3x – 5)9 . Or you can use your calculator and enter d ((3x – 5)10 , x ) and obtain the same result.

Example 2

If , find f ′(x ).

Rewrite as f (x ) = 5x (25 – x 2 )1/2. Using the product rule, f ′(x ) = .

To find , use the chain rule and let u = 25 – x 2 .

Thus, . Substituting this quantity back into f ′(x ), you have . Or you can use your calculator and enter and obtain the same result.

Example 3

If , find .

Using the chain rule, let . Then .

To find , use the quotient rule.

Thus, . Substituting this quantity back into .

An alternate solution is to use the product rule and rewrite as y = and use the quotient rule. Another approach is to express y = (2x – 1)3 (x –6 ) and use the product rule. Of course, you can always use your calculator if you are permitted to do so.

7.2 Derivatives of Trigonometric, Inverse Trigonometric, Exponential, and Logarithmic Functions

Main Concepts: Derivatives of Trigonometric Functions, Derivatives of Inverse Trigonometric Functions, Derivatives of Exponential and Logarithmic Functions

Derivatives of Trigonometric Functions
Summary of Derivatives of Trigonometric Functions

Note that the derivatives of cosine , cotangent , and cosecant all have a negative sign.

Example 1

If y = 6x 2 + 3 sec x , find .

sec x tan x .

Example 2

Find f ′(x ) if f (x ) = cot(4x – 6).

Using the chain rule, let u = 4x – 6. Then f ′(x ) = [– csc2 (4x – 6)][4] = –4 csc2 (4x – 6).

Or using your calculator, enter d (1/tan(4x – 6), x ) and obtain which is an equivalent form.

Example 3

Find f ′(x ) if f (x ) = 8 sin(x 2 ).

Using the chain rule, let u = x 2 . Then f ′(x ) = [8 cos(x 2 )][2x ] = 16x cos(x 2 ).

Example 4

If y = sin x cos(2x ), find .

Using the product rule, let u = sin x and v = cos(2x ).

Then .

Example 5

If y = sin[cos(2x )], find .

Using the chain rule, let u = cos(2x ). Then

.

To evaluate , use the chain rule again by making another u -substitution, this time for 2x . Thus, . Therefore, .

Example 6

Find f ′(x ) if f (x ) = 5x csc x .

Using the product rule, let u = 5x and v = csc x . Then f ′(x ) = 5 csc x + (– csc x cot x ) (5x ) = 5 csc x – 5x (csc x )(cot x ).

Example 7

If , find .

Rewrite as y = (sin x ) 1/2 . Using the chain rule, let u = sin x . Thus, .

Example 8

If , find .

Using the quotient rule, let u = tan x and v = (1 + tan x ). Then,

Note: For all of the above exercises, you can find the derivatives by using a calculator, provided that you are permitted to do so.

Derivatives of Inverse Trigonometric Functions
Summary of Derivatives of Inverse Trigonometric Functions

Let u be a differentiable function of x , then

Note that the derivatives of cos–1 x , cot–1 x , and csc–1 x all have a “–1” in their numerators.

Example 1

If y = 5 sin–1 (3x ), find .

Let u = 3x . Then .

Or using a calculator, enter d [5 sin–1 (3x ), x ] and obtain the same result.

Example 2

Find f ′(x ) .

Let

Example 3

If y = sec–1 (3x 2 ), find .

Let u = 3x 2 . Then .

Example 4

If .

Note: For all of the above exercises, you can find the derivatives by using a calculator, provided that you are permitted to do so.

Derivatives of Exponential and Logarithmic Functions
Summary of Derivatives of Exponential and Logarithmic Functions

Let u be a differentiable function of x , then

.

For the following examples, find and verify your result with a calculator.

Example 1

y = e 3x + 5xe 3 + e 3

(Note that e 3 is a constant.)

Example 2

y = xex x 2 ex

Using the product rule for both terms, you have

Example 3

y = 3sin x

Let u = sin x . Then, .

Example 4

y = e (x 3 )

Let u = x 3 . Then, .

Example 5

y = (ln x )5

Let u = ln x . Then, .

Example 6

y = ln(x 2 + 2x – 3) + ln 5

Let u = x 2 + 2x – 3. Then, .

(Note that ln 5 is a constant. Thus the derivative of ln 5 is 0.)

Example 7

y = 2x ln x + x

Using the product rule for the first term,

you have (2x ) + 1 = 2 ln x + 2 + 1 = 2 ln x + 3.

Example 8

y = ln(ln x )

Let u = ln x . Then .

Example 9

y = log5 (2x + 1)

Let u = 2x + 1. Then .

Example 10

Write an equation of the line tangent to the curve of y = ex at x = 1.

The slope of the tangent to the curve y = ex at x = 1 is equivalent to the value of the derivative of y = e x evaluated at x = 1. Using your calculator, enter d (e (x ), x )|x = 1 and obtain e . Thus, m = e , the slope of the tangent to the curve at x = 1. At x = 1, y = e 1 = e , and thus the point on the curve is (1, e ). Therefore, the equation of the tangent is ye = e (x – 1) or y = ex . (See Figure 7.2-1 .)

Figure 7.2-1

• Never leave a multiple-choice question blank. There is no penalty for incorrect answers.

7.3 Implicit Differentiation

Main Concept: Procedure for Implicit Differentiation

Procedure for Implicit Differentiation

Given an equation containing the variables x and y for which you cannot easily solve for y in terms of x , you can find by doing the following:

Steps

1: Differentiate each term of the equation with respect to x .

2: Move all terms containing to the left side of the equation and all other terms to the right side.

3: Factor out on the left side of the equation.

4: Solve for .

Example 1

Find if y 2 – 7y + x 2 – 4x = 10.

Step 1: Differentiate each term of the equation with respect to x . (Note that y is treated as a function of x .)

Step 2: Move all terms containing to the left side of the equation and all other terms to the right: .

Step 3: Factor out .

Step 4: Solve for .

Example 2

Given x 3 + y 3 = 6xy , find .

Step 1: Differentiate each term with respect to .

Step 2: Move all terms to the left side: .

Step 3: Factor out .

Step 4: Solve for .

Example 3

Find if (x + y ) 2 – (xy ) 2 = x 5 + y 5 .

Step 1: Differentiate each term with respect to x :

Distributing 2(x + y ) and –2(xy ), you have.

Step 2: Move all terms to the left side:

Step 3: Factor out :

Step 4: Solve for .

Example 4

Write an equation of the tangent to the curve x 2 + y 2 + 19 = 2x + 12y at (4, 3). The slope of the tangent to the curve at (4, 3) is equivalent to the derivative at (4, 3).

Using implicit differentiation, you have:

Thus, the equation of the tangent is y – 3 = (1)(x – 4) or y – 3 = x – 4.

Example 5

Find , if sin(x + y ) = 2x .

.

7.4 Approximating a Derivative

Given a continuous and differentiable function, you can find the approximate value of a derivative at a given point numerically. Here are two examples.

Example 1

The graph of a function f on [0, 5] is shown in Figure 7.4-1 . Find the approximate value of f ′(3).

Figure 7.4-1

Since f ′(3) is equivalent to the slope of the tangent to f (x ) at x = 3, there are several ways you can find its approximate value.

Method 1: Use the slope of the line segment joining the points at x = 3 and x = 4.

Method 2: Use the slope of the line segment joining the points at x = 2 and x = 3.

Method 3: Use the slope of the line segment joining the points at x = 2 and x = 4.

Note that is the average of the results from methods 1 and 2.

Thus, f ′(3) ≈ 1, 2, or depending on which line segment you use.

Example 2

Let f be a continuous and differentiable function. Selected values of f are shown below. Find the approximate value of f ′ at x = 1.

You can use the difference quotient to approximate f ′(a ).

Or, you can use the symmetric difference quotient to approximate f ′(a ).

Thus, f ′(3) ≈ 0.49, 0.465, 0.54, or 0.63 depending on your method.

Note that f is decreasing on (–2, –1) and increasing on (–1, 3). Using the symmetric difference quotient with h = 3 would not be accurate. (See Figure 7.4-2 .)

Figure 7.4-2

• Remember that the because the .

7.5 Derivatives of Inverse Functions

Let f be a one-to-one differentiable function with inverse function f –1 . If f ′(f –1 (a )) ≠ 0, then the inverse function f –1 is differentiable at a and (f –1 )′(a ) = . (See Figure 7.5-1 .)

Figure 7.5-1

If y = f –1 (x ) so that x = f (y ), then with .

Example 1

If f (x ) = x 3 + 2x – 10, find (f –1 )′(x ).

Step 1: Check if (f –1 )′(x ) exists. f ′(x ) = 3x 2 + 2 and f ′(x ) > 0 for all real values of x . Thus, f (x ) is strictly increasing which implies that f (x ) is 1 – 1. Therefore, (f –1 )′(x ) exists.

Step 2: Let y = f (x ) and thus y = x 3 + 2x – 10.

Step 3: Interchange x and y to obtain the inverse function x = y 3 + 2y – 10.

Step 4: Differentiate with respect to .

Step 5: Apply formula

Example 2

Example 1 could have been done by using implicit differentiation.

Step 1: Let y = f (x ), and thus y = x 3 + 2x – 10.

Step 2: Interchange x and y to obtain the inverse function x = y 3 + 2y – 10.

Step 3: Differentiate each term implicitly with respect to x .

Step 4: Solve for .

Example 3

If f (x ) = 2x 5 + x 3 + 1, find (a) f ′(1) and f ′(1) and (b) (f –1 )(4) and (f –1 )′(4).

Enter y 1 = 2x 5 + x 3 + 1. Since y 1 is strictly increasing, f (x ) has an inverse.

(a) f (1) = 2(1)5 + (1)3 + 1 = 4
f ′(x ) = 10x 4 + 3x 2
f ′(1) = 10(1)4 + 3(1)2 = 13

(b) Since f (1) = 4 implies the point (1, 4) is on the curve f (x ) = 2x 5 + x 3 + 1, therefore, the point (4, 1) (which is the reflection of (1, 4) on y = x ) is on the curve (f –1 )(x ). Thus, (f –1 )(4) = 1.

Example 4

If f (x ) = 5x 3 + x + 8, find (f –1 )′(8).

Enter y 1 = 5x 3 + x + 8. Since y 1 is strictly increasing near x = 8, f (x ) has an inverse near x = 8.

Note that f (0) = 5(0)3 + 0 + 8 = 8 which implies the point (0, 8) is on the curve of f (x ).

Thus, the point (8, 0) is on the curve of (f –1 )(x ).

Therefore,

• You do not have to answer every question correctly to get a 5 on the AP Calculus AB exam. But always select an answer to a multiple-choice question. There is no penalty for incorrect answers.

7.6 Higher Order Derivatives

If the derivative f ′ of a function f is differentiable, then the derivative of f ′ is the second derivative of f represented by f ″ (reads as f double prime). You can continue to differentiate f as long as there is differentiability.

Some of the Symbols of Higher Order Derivatives

Note that .

Example 1

If y = 5x 3 + 7x – 10, find the first four derivatives.

Example 2

If , find f ″ (4).

Rewrite: and differentiate: .

Differentiate again:

Example 3

If y = x cos x , find y ″.

Or, you can use a calculator and enter d [x cos x , x , 2] and obtain the same result.

7.7 L’Hôpital’s Rule for Indeterminate Forms

Let lim represent one of the limits: or . Suppose f (x ) and g (x ) are differentiable and g ′(x ) = 0 near c , except possibly at c , and suppose lim f (x ) = 0 and lim g (x ) = 0. Then the is an indeterminate form of the type . Also, if lim f (x ) = ±∞ and lim g (x ) = ±∞, then the , is an indeterminate form of the type . In both cases, and , L’Hôpital’s Rule states that lim = lim .

Example 1

Find lim , if it exists.

Since , this limit is an indeterminate form. Taking the derivatives, and . By L’Hôpital’s Rule, .

Example 2

Find , if it exists.

Rewriting as shows that the limit is an indeterminate form, since and . Differentiating and applying L’Hôpital’s Rule means that . Unfortunately, this new limit is also indeterminate. However, it is possible to apply L’Hôpital’s Rule again, so equals to . This expression approaches zero as x becomes large, so .

7.8 Rapid Review

1. If y = ex 3 , find .

Answer: Using the chain rule, .

2. Evaluate .

Answer: The limit is equivalent to .

3. Find f ′(x ) if f (x ) = ln(3x ).

4. Find the approximate value of f ′(3). (See Figure 7.8-1 .)

Figure 7.8-1

Answer: Using the slope of the line segment joining (2, 1) and (4, 3), .

5. Find if xy = 5x 2 .

Answer: Using implicit differentiation, . Thus, .

Or simply solve for y leading to y = 5x and thus, .

6. If , find .

Answer: Rewrite y = 5x - 2. Then, = and .

7. Using a calculator, write an equation of the line tangent to the graph f (x ) = –2x 4 at the point where f ′(x ) = –1.

Answer: f ′(x ) = –8x 3 . Using a calculator, enter [Solve ] [–8x 3 =-1, x ] and obtain . Using the calculator . Thus, tangent is .

8.

9.

7.9 Practice Problems

Part A The use of a calculator is not allowed.

Find the derivative of each of the following functions.

1 . y = 6x 5x + 10

2 .

3 .

4 .

5 . f (x ) = (3x – 2)5 (x 2 – 1)

6 .

7 . y = 10 cot(2x – 1)

8 . y = 3x sec(3x )

9 . y = 10 cos[sin(x 2 – 4)]

10 . y = 8 cos–1 (2x )

11 . y = 3e 5 + 4xe x

12 . y = ln(x 2 + 3)

Part B Calculators are allowed.

13 . Find , if x 2 + y 3 = 10 – 5xy .

14 . The graph of a function f on [1, 5] is shown in Figure 7.9-1 . Find the approximate value of f ′(4).

Figure 7.9-1

15 . Let f be a continuous and differentiable function. Selected values of f are shown below. Find the approximate value of f ′ at x = 2.

16 . If f (x ) = x 5 + 3x – 8, find (f –1 )′(–8).

17 . Write an equation of the tangent to the curve y = ln x at x = e .

18 . If y = 2x sin x , find at .

19 . If the function f (x ) = (x – 1)2 /3 + 2, find all points where f is not differentiable.

20 . Write an equation of the normal line to the curve x cos y = 1 at .

21 .

22 .

23 .

24 .

25 .

7.10 Cumulative Review Problems

(Calculator) indicates that calculators are permitted.

26 . Find

27 . If f (x ) = cos2 (πx ), find f ′(0).

28 . Find .

29 . (Calculator) Let f be a continuous and differentiable function. Selected values of f are shown below. Find the approximate value of f ′ at x = 2.

30 . (Calculator) If determine if f (x ) is continuous at (x = 3). Explain why or why not.

7.11 Solutions to Practice Problems

Part A The use of a calculator is not allowed.

1 . Applying the power rule, .

2 . Rewrite as f (x ) = x –1 + x –2/3 . Differentiate: .

3 . Rewrite .
Differentiate: .
An alternate method is to differentiate directly, using the quotient rule.

4 . Applying the quotient rule,

5 . Applying the product rule, u = (3x – 2)5 and v = (x 2 – 1), and then the chain rule,

6 . Rewrite as.

Applying first the chain rule and then the quotient rule,

Note: ,

if which implies or .

Another method is to write and use the product rule.

Another method is to write y = (2x + 1)1/2 (2x – 1)1/2 and use the product rule.

7 . Let u = 2x – 1,

8 . Using the product rule,

9 . Using the chain rule, let u = sin(x 2 – 4).

10 . Using the chain rule, let u = 2x .

11 . Since 3e 5 is a constant, its derivative is 0.

12 . Let

Part B Calculators are allowed.

13 . Using implicit differentiation, differentiate each term with respect to x .

14 . Since f ′(4) is equivalent to the slope of the tangent to f (x ) at x = 4, there are several ways you can find its approximate value.

Method 1: Use the slope of the line segment joining the points at x = 4 and x = 5.

Method 2: Use the slope of the line segment joining the points at x = 3 and x = 4.

Method 3: Use the slope of the line segment joining the points at x = 3 and x = 5.

Note that –2 is the average of the results from methods 1 and 2. Thus f ′(4) ≈ –3, –1, or –2 depending on which line segment you use.

15 . You can use the difference quotient to approximate f ′(a ). Let h = 1; .

Or, you can use the symmetric difference quotient to approximate f ′(a ).

Let h = 1; .

Thus, f ′(2) ≈ 4 or 5 depending on your method.

16 . Enter y 1 = x 5 + 3x – 8. The graph of y 1 is strictly increasing. Thus f (x ) has an inverse. Note that f (0) = –8. Thus the point (0, –8) is on the graph of f (x ) which implies that the point (–8, 0) is on the graph of f 1(x ).
f ′(x ) = 5x 4 + 3 and f ′(0) = 3.

Since , thus .

17 . and Thus the slope of the tangent to y = ln x at x = e is . At x = e , y = ln x = ln e = 1, which means the point (e ,1) is on the curve of y = ln x . Therefore, an equation of the tangent is or See Figure 7.11-1 .

Figure 7.11-1

18 .

Or, using a calculator, enter d (2x – sin(x ), x , 2) and obtain –π .

19 . Enter y 1 = (x – 1)2 /3 + 2 in your calculator. The graph of y 1 forms a cusp at x = 1. Therefore, f is not differentiable at x = 1.

20 . Differentiate with respect to x :

Thus, the slope of the tangent to the curve at (2, π /3) is The slope of the normal line to the curve at (2, π /3) is . Therefore an equation of the normal line is .

21 .

22 .

23 .

24 .

25 .

7.12 Solutions to Cumulative Review Problems

26 . The expression the derivative of sin x at x = π /2 which is the slope of the tangent to sin x at x = π /2. The tangent to sin x at x = π /2 is parallel to the x -axis.
Therefore the slope is 0, i.e., . An alternate method is to expand sin .

27 . Using the chain rule, let u = (πx ).

Then,

28 . Since the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator, the limit is 0.

29 . You can use the differencequotient to approximate f ′(a ).

Or, you can use the symmetric difference quotient to approximate f ′(a ).

Thus, f ′(2) = 1.7, 2.05, or 1.25 depending on your method.

30 . (See Figure 7.12–1 .) Checking the three conditions of continuity:

Figure 7.12-1

(1) f (3) = 3

(2)

(3) Since , f (x ) is discontinuous at x = 3.

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