Find the Extrema:

þ                                         ¦ (x)= 3x⁴ –4x³ on interval [-1,2]

þ                                         For Critical Numbers get 1^st derivative and set to zero

þ                                         ¦’(x) = 12x³-12x² 

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

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

þ                                         ¦(-1)= 3x⁴ –4x³  =   7 Þ  (-1,7)

þ                                         ¦(0)= 3x⁴ –4x³  =   0 Þ   (0,0)

þ                                         ¦(1)= 3x⁴ –4x³  =   -1 Þ  (1, -1)Þ Minimum

þ                                         ¦(2)= 3x⁴ –4x³  =   16 Þ  (2,16)Þ Maximum

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

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) ß(This part makes it water unevenly.)

(Cos2q) ß(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)] ß(This part is the constant)

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

Section 3.2 (Rolle’s Theorem & the Mean Value Theorem)

*** Key things to remember:

Rolle’s Theorem & Mean Value Theorem (MVT).

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

þ                 For Rolle’s 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.

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

To work a problem, here is an example:

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 Rolle’s 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 Þ 4 minutes or 5 miles Þ 50MPH

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 1^st 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 < x_2,

        and ¦’(c)= [¦(x₂)- ¦(x₁)]/[x_{2 -} x₁].

• Because ¦’(c)>0 and x_{2 -} 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 é or ê.

Let f be continuous on the intervals (a,b). To find the open intervals on which ¦ is é or ê, 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 é or ê 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:

þ                 F (x) = x³ - 12x (need 1^st derv for critical points)

þ                 F’ (x) = 3x² - 12 

þ                 Set to zero 8 0 = 3x² - 12 and solve.

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

þ                  

Theorem 3.6  - The 1^st 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 ¦.

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

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

Relative minimum È, Relative maximum Ç.

If like a sideways “S” then neither Min nor Max.

Definition of concavity

Let ¦ be differentiable on an open interval I. 

The graph of ¦ is Concave UP È on I if ¦’ is increasing é on the interval and Concave DOWN Ç on I if ¦’is Decreasing ê on the interval.

Another example of Determining Concavity:

¦ (x)= 6/(x² + 3)« 6(x² + 3)^-1

¦’(x)= (-6)( x² + 3)^-2 (2x) « -12x/(x² + 3)²

¦”(x)= [(x² + 3)²(-12)-(-12x)(2)( x² + 3)(2x)]/( x² + 3)³

      = [36(x² +-1)]/(x² + 3)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:

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. 

þ                  ¦ (x)= 3x⁴ –4x³ on interval [-1,2]

þ                 For Critical Numbers get 1^st derivative and set to zero

þ                 ¦’(x) = 12x³-12x² 

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

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

þ                 ¦(-1)= 3x⁴ –4x³  = 7 Þ (-1,7)

þ                 ¦(0)= 3x⁴ –4x³  = 0 Þ             (0,0)

þ                 ¦(1)= 3x⁴ –4x³  = -1 Þ           (1, -1)Þ Minimum

þ                 ¦(2)= 3x⁴ –4x³  = 16 Þ           (2,16)Þ Maximum

þ                 Now we look at derivative of ¦’(x) = 12x³-12x² which is the second derivative of ¦(x)= 3x⁴ –4x³

þ                 ¦” (x)= 12x² – 24 « 12x(x –2)

þ                 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.

þ                 Here is the Table:

þ                  

Using the 2^nd Derivative Test.

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

Another example using the 2^nd Derivative Test.

Find the relative extrema for ¦ (x) = -3x⁵ + 5x³

Solution: Begin by finding the critical numbers of ¦.

¦′(c)= -15x⁴ + 15x² « 15x²(1 - x²)= 0

Critical numbers:  x=-1,0,1

Using the 2^nd Der: ¦”(c)= -60x³ + 30x « = 30(-2x³ + x)

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

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.

^ü        Example:  N ⁴ / Q ³

^ü        The limit is Undefined or it Does Not Exist.

^ü        No horizontal asymptote.

Horizontal Asymptote:  

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

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

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

A Comparison of 3 Rational Functions

¦(x)=(2x + 5)/(3x² + 1){Divide both the N & Q by x²} « 0/3 = 0

¦(x)=(2x² + 5)/(3x² + 1){Divide both the N & Q by x²} « 2/3

¦(x)=(2x³ + 5)/(3x² + 1){Divide both the N & Q by x²} « ¥/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)/(2x² + 1)^1/2 « H.A.= 3 /(2)^1/2  

For x < 0

¦(x)= (3x-2)/(2x² + 1)^1/2 « H.A.= 3 /-(2)^1/2  

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:

þ                 Y = 3x⁴ + 4x³ « 0= x³(3s + 4)«X-Int:  x = 0, & x = -4/3

þ                 Y’= 12x³ + 12x « 0= x²(x + 1)«Critical Points: x = 0, &  x = -1

þ                 Y” = 36x² + 24x « 0= 12x(3x + 2)«Test Values: x = 0, & x = -2/3 

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 - 14x² + 4x³

V'(x)= 12 + 28x + 12x² " 0 =12+ 28x + 12x²

    0=4(3 + 7x + 3x²) " 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

þ                 See special section on "Computer Stuff Page"

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

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

þ                 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).

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

Formula:   

X_(n+1)= Xn - ¦ (X_n) ¬(this is the original equation)                  

           ¦′(X_n) ¬(this is the 1st derivative of that equation