Chapter 3 notes for Calc 1

Section 3.1 (Extrema on a Interval)

Find the Extrema:

                                         (x)= 3x4 4x on interval [-1,2]

                                         For Critical Numbers get 1st derivative and set to zero

                                         (x) = 12x-12x 

                                         0= 12x(x-1)     therefore: x=0,x=1 (Critical Numbers)

                                         Now consider all point, the Critical and the End points.

                                         (-1)= 3x4 4x  =   7   (-1,7)

                                         (0)= 3x4 4x  =   0    (0,0)

                                         (1)= 3x4 4x  =   -1   (1, -1) Minimum

                                         (2)= 3x4 4x  =   16   (2,16) Maximum

 

*** To use TI-86 to get Min/Max

Graph: Enter Function Exit Exit 2nd, calc, more,

Select either the Min or Max, you will see like this Min(

Here you enter y1 (get from catalog), Min(y1

Enter a coma, then variable x, Then a Minimum number,

Then a Maximum number, Then close parentheses)

**** It will look like this Min (y1,x,0,325)

Enter, (this gives the x value).  Sto x, (store it in x)

Now you enter y1 (get from catalog),

Hit enter:(this is (x))

   

Problem 56 on page 167 of text book:

Lawn Sprinkler: A lawn sprinkler is constructed in such a way that dq/dt is constant, where q ranges between 45 and 135.  The distance the water travels horizontally is  

   x = (v2 sinq)/32,    45 q 135 

Where v is the speed of the water.  Find dx/dt and explain why this lawn sprinkler does not water evenly.  What part of the lawn receives the most water? 

Solution: 

Find dx/dt = ((v2)/32)* (cos2q)* 2 (dq/dt)

dx/dt=[((v2)/16)*(dq/dt)](cos2q)  

**********************

** Special notes:

(Cos2q) (This part makes it water unevenly.)

(Cos2q) (This part is close to zero at 135 and 45.

Therefore, most watering occurs at these 2 points    with less watering in-between).

[((v2)/16)*(dq/dt)] (This part is the constant)

 

***Continue problem solution: Need to finish!!!

Section 3.2 (Rolles Theorem & the Mean Value Theorem)

*** Key things to remember:

Rolles Theorem & Mean Value Theorem (MVT).

                 Cannot have sharp point (like that of |x|), but can have a smooth curve.

                 For Rolles Theorem: The function must have some point where the derivative = zero (at least 1 of more).  If not at least 1 point equal to zero, then it is not continuous and not differentiable.

                 ** Sin, Cos, and all polynomials are all differentiable and continuous, everywhere.

To work a problem, here is an example:

(x)= 7x4 - 24x3 + 10x2 + 24x 17

For (x):  Get all x intercepts, y intercepts and points = 0 

Set (x)= 0

0 = 7x4 - 24x3 + 10x2 + 24x - 17 (x - 1) 2 (7x - 17) (x + 1)

X-Intercepts: (-1,0) (1,0) ((17/7), 0) (here y=0)

Y-Intercepts: (0,17) (here x=0 & all cancel except 17)

   

(x)= (4)(7) x3 -(3)(24) x2 + (2)(10) x + 24 

          = 28x3 72x2 + 20x + 24

           = 4(7 x3 - 18x2  + 5x + 8)         

           = 4(x - 1)(x 2)(7x + 3)

Therefore, the x roots of (x):

x = 1, x = -3/7, x = 2

Now check the x roots of (x):

by plugging them into (x)= 7x4 - 24x3 + 10x2 + 24x 17

 (1)=0,

(2)= -9,

(-3/7)= -23.3236

Therefore, the critical points are:    (1, 0), (2, -9), (-3/7, -23.3236)

(1, 0) is a relative Max point.

(2, -9) is a relative Min point.

(-3/7, -23.3236) is a relative min point.

 

Proof of MVT:

The equation of secant line containing the point (a,f(a)) and (b,f(b)) is

Y= [ (f(b) f(a))/ (b-a)] *(x-a) + f(a).

 

Let g(x) be the difference between f(x) and y, then

  g(x) = f(x) y = f(x) - [ (f(b) f(a))/ (b-a)] *(x-a) - f(a).

 

By evaluating g at a and b, you can see that g(a)=0=g(b).

Furthermore, because f is differentiable, g is also differentiable, and you can apply Rolles Theorem to the function g.

So, there exists a number c in (a,b) such that g(0), which implies that

0 = g(c )

   = f (c ) [ (f(b) f(c))/ (b a) ]

 

Therefore, there exists a number c in (a,b) such that

   f (c ) = [ (f(b) f(c))/ (b a) ]

 

Word Problem:

Car & Speed trap:

Start point 55MPH 4 minutes or 5 miles 50MPH

D = rt

5 = r (4min/60min)

300 = 4r

r = 75 MPH

Average speed given that distance and that time. 

D (t)= distance in terms of time.

D (0) = o miles

D (4 min)= 5 miles

D (1/15 hr)= 5 miles

[(b)- (a)]/[(b)-(a)]

[(1/15)- (0)]/[(1/15)-0]=75MPH

Section 3.3 (Increasing & Decreasing F(x) & the 1st Derivative Test)

Theorem 3.5 _ Test for Increasing & Decreasing Functions

Let be a function that is continuous on the closed interval [a,b] and the differentiable on the open interval(a,b).

 

  1. If (x) > 0 for all x in (a,b), then is increasing on [a,b].
  2. If (x) < 0 for all x in (a,b), then is decreasing on [a,b].
  3. If (x) = 0 for all x in (a,b), then is constant on [a,b].

 

Proof:

  • To prove the 1st case, assume that (x) > 0 for all x in the interval (a,b) and

      let x1 < x2 be any 2 points in the interval. 

  • By the MVT, you know that there exists a number c such that x1 < c < x2,

        and (c)= [(x2)- (x1)]/[x2 - x1].

  • Because (c)>0 and x2 - x1 > 0, you know that (x2)-f(x1)> 0

            which implies that (x1) < (x2). So, is increasing on the interval.

 

Guidelines for Finding Intervals on Which a Function is or .

Let f be continuous on the intervals (a,b). To find the open intervals on which is or , use the fallowing steps.

1.      Locate the critical numbers of f in (a,b), & use these numbers to determine test intervals.

2.      Determine the sign of (x) at 1 test value for each of the intervals.

3.      Use Theorem 3.5 to determine whether f is or on each interval.

 

These guidelines are also valid if the interval (a,b) is replaced by an interval of the form (-,b)(a,),or (-,).

 

For implementation of these steps, see next section.

 

See last example on Exam 3 web page for sketch of this function:

                 F (x) = x - 12x (need 1st derv for critical points)

                 F (x) = 3x - 12 

                 Set to zero 8 0 = 3x - 12 and solve.

                 X = -2 and X = 2, therefore -2 < x < 2 and set interval to evaluate if or .

                  

Interval

- < x < -2

-2 < x < 2

2 < x <

Test value

- 3

0

3

Sign of (x)

(-3) =15 > 0

(0) = -12 < 0

(3) =15 > 0

Conclusion

Increasing

Decreasing

Increasing

 

 

Theorem 3.6  - The 1st Derivative Test

 

Let c be a critical number of a function that is continuous on and open interval I containing c.  If is differentiable on the interval, except possibly at the c, then (c) can be classified as follows.

 

  1. If (x) changes from negative to positive at c, then (c) is a relative minimum of .

 

  1. If (x) changes from positive to negative at c, then f(c) is a relative maximum of .

 

  1. If (x) does not change sign at c, then (c) is neither a relative minimum nor a relative maximum.

 

Relative minimum , Relative maximum .

If like a sideways S then neither Min nor Max.

 

Section 3.4 (Concavity & the 2nd Derivative Test)

 

Definition of concavity


Let be differentiable on an open interval I. 

The graph of is Concave UP on I if is increasing on the interval and Concave DOWN on I if is Decreasing on the interval.

 

 

We will look at a function from earlier and see how it is evaluated for Concavity. 

 

The function is: (x)= 7x4 - 24x3 + 10x2 + 24x 17

The 1st Derivative:(x)= 28x3 72x2 + 20x + 24

The 2nd Derivative = 84x2 144x + 20 4(21 x2 36x + 5)

 

Where it will be zero for Concave UP or Concave DOWN .

 

4(21 x2 36x + 5) = 0 Quadratic formula x= (36 +or- (362 4(21)(5)))/ 42

x=0.1524453 and  x=1.56184 now set these as intervals.

This is a table of the intervals

(-, 0.1524),

(0.152, 1.56184)

(1.56184, )

((-3)=Pos)

((1)=Neg)

((5)=Pos)

Concave UP 

Concave DOWN  ,

Concave UP

 

 

 

 

 

Another example of Determining Concavity:

 

(x)= 6/(x + 3) 6(x + 3)-1

 

(x)= (-6)( x + 3)-2 (2x) -12x/(x + 3)

 

(x)= [(x + 3)(-12)-(-12x)(2)( x + 3)(2x)]/( x + 3)

      = [36(x +-1)]/(x + 3)

 

Because (x) = 0 when x = 1 and is defined on the entire real line, you should test on the intervals (-, -1), (-1,1), and (1,).  The results are shown in the table:

(-, -1),

(-1, 1)

(1, )

(-2) > 0

(0) < 0

((2) > 0

Concave UP 

Concave DOWN  ,

Concave UP

 

 

 

 

Points of Inflection

The Concavity of changes at a point of inflection.

 

THEOREM 3.8 - Points of Inflection

If (c, (c)) is a point of inflection on the graph of , then either (c)= 0 pr is not differentiable at x = c. 

 

 

                   (x)= 3x4 4x on interval [-1,2]

                 For Critical Numbers get 1st derivative and set to zero

                 (x) = 12x-12x 

                 0= 12x(x-1)     therefore: x=0,x=1 (Critical Numbers)

                 Now consider all point, the Critical and the End points.

                 (-1)= 3x4 4x  = 7 (-1,7)

                 (0)= 3x4 4x  = 0             (0,0)

                 (1)= 3x4 4x  = -1            (1, -1) Minimum

                 (2)= 3x4 4x  = 16            (2,16) Maximum

                 Now we look at derivative of (x) = 12x-12x which is the second derivative of (x)= 3x4 4x

                 (x)= 12x 24 12x(x 2)

                 Set (x) = 0, you can determine that the possible points of inflection occur at x=0 and x=2.  By testing the intervals determined by these x- values, you can conclude that they both yield points of inflection.

                 Here is the Table:

 

                  

Interval

- < x < 0

0 < x < 2

2 < x <

Test value

X = - 1

X = 1

X = 3

Sign of (x)

(-1) > 0

(1) < 0

(3) > 0

Conclusion

Concave UP 

Concave DOWN  ,

Concave UP 

 

Using the 2nd Derivative Test.

In addition to testing for concavity, the 2nd Derivative Test can be used to perform a simple test for relative maxima & minima.

THEOREM 3.9 Second Derivative Test

Let be a function such that (c)= 0 and the 2nd derivative of exists on an open interval containing c.

1.              If (c)> 0, then (c) is a relative minimum.

2.              If (c)< 0, then (c) is a relative maximum.

3.              If (c)= 0, the test fails.  In such cases, you use the 1st Derivative Test.

 

 

Another example using the 2nd Derivative Test.

Find the relative extrema for (x) = -3x5 + 5x

Solution: Begin by finding the critical numbers of .

 

(c)= -15x4 + 15x 15x(1 - x)= 0

Critical numbers:  x=-1,0,1

Using the 2nd Der: (c)= -60x + 30x = 30(-2x + x)

 

Point

(-1, -2)

(1,2)

(0,0)

Sign of (x)

(-1) > 0

(1) < 0

(0) = 0

Conclusion

Relative minimum

Relative maximum

Test fails

 

Because the 2nd Derivative Test fails at (0,0), you can use the 1st Derivative Test and observe that to the left and the right of x=0.  SO, 90,0) is neither a relative min nor a relative max.

 

 

Section 3.5  (Limits at Infinity)

Useful things to review for this section:  

Let the numerator be represented by N and the Denominator be represented by Q, then you may use the formulas below. 

 

Vertical Asymptote:

Set Q(c)=0, Then x = c

Slant Asymptote:

Exists ONLY if the degree of N is EXACTLY 1 more than the degree of Q.

        Example:  N 4 / Q 3

        The limit is Undefined or it Does Not Exist.

        No horizontal asymptote.

 

Horizontal Asymptote:  

                 N < Q, then y = the x-axis (The limit is 0 & y = 0) 

                 N = Q, then y = the Leading Coefficients of N/Q.  

                 N > Q, then the horizontal asymptote Does Not Exist.

 

A Comparison of 3 Rational Functions

(x)=(2x + 5)/(3x + 1){Divide both the N & Q by x} 0/3 = 0

(x)=(2x + 5)/(3x + 1){Divide both the N & Q by x} 2/3

(x)=(2x + 5)/(3x + 1){Divide both the N & Q by x} /3 = DNE

Guidelines for finding the Limits at Infinity of Rational Functions

  1. If the degree of the numerator is less than the degree of the denominator, then the limit of the rational function is 0.
  2. If the degree of the numerator is equal to the degree of the denominator, then the limit of the rational function is the ratio of the leading coefficients.
  3. If the degree of the numerator is greater than the degree of the denominator, then the limit of the rational function does not exist.

 

An Example with 2 Horizontal Asymptote:

For x > 0

(x)= (3x-2)/(2x + 1)1/2 H.A.= 3 /(2)1/2  

For x < 0

(x)= (3x-2)/(2x + 1)1/2 H.A.= 3 /-(2)1/2  

 

Section 3.6  (A summary of Curve Sketching)

Guidelines for Analyzing the Graph of a Function:

 

1.      Determine the domain and range of the function.

2.      Determine the intercepts, asymptotes, & symmetry of the graph.

3.      Locate the x-values for which ′(x) & ′′ (x) are either zero or do not exist.  Use the results to determine relative extrema & points of inflection.

 

An Example of using all the information we have learned:

                 Y = 3x4 + 4x 0= x(3s + 4)X-Int:  x = 0, & x = -4/3

                 Y= 12x + 12x 0= x(x + 1)Critical Points: x = 0, &  x = -1

                 Y = 36x + 24x 0= 12x(3x + 2)Test Values: x = 0, & x = -2/3 

 

 

(x)

′(x)

(x)

Characteristic of Graph

- < x < -1

 

-

+

Decreasing, Concave Up

X = -1

-1

0

+

Relative minimum

-1 < x < -2/3

 

+

+

Increasing, Concave Up

X = -2/3

-16/27

+

0

Point of inflection

-2/3 < x < 0

 

+

-

Increasing, Concave Down

X = 0

0

0

0

Point of inflection

0 < x <

 

+

+

Increasing, Concave Up

 

 

 

 

Section 3.7  (Optimization Test)

 

Word Problem:  Make a box with open top. 

What is the size of the box of volume is Maximized?

** The base of box along side measuring 3': 3'-2x

** The base of box along side measuring 4': 4'-2x

** The height of the box: x

Volume is V(x)=x(3-2x)(4-2x)

The limits are: has to be less than 3/2 therefore 0 < x < 3/2

V(x)=x(3-2x)(4-2x)  "  12x - 14x + 4x

V'(x)= 12 + 28x + 12x " 0 =12+ 28x + 12x

    0=4(3 + 7x + 3x) " x=(7(13)1/2)/6 

Since x=(7+(13)1/2)/6 = 1.76759 it is > 3/2 we cannot use this             

but x=(7-(13) 1/2)/6 = 0.565741454089

x is approximately 0.5657

 

Conclusion:

The approximate dimensions of the box:

Height is about 0.5657,

Width is 3-2(.5657) " 1.8686

Length is 4-2(.5657) " 2.8686

 

 

Section 3.8  (Newtons Method)

 

                 See special section on "Computer Stuff Page"

                 Best link for doing a Complete Calculation including the 1st Derivative:

                 http://mss.math.vanderbilt.edu/~pscrooke/MSS/newtonnum.html

                 The above link starts with the function, will determine the 1st derivative, then will give all the results of each calculation until it reaches the end point (the Root).

                 You will need to give the 1st Derivative as a step in your answer.

Formula:   

X(n+1)= Xn - (Xn) (this is the original equation)                  

           (Xn) (this is the 1st derivative of that equation