College Algebra to Calculus and the T-83
Lesson 23
More on maximum and minimum
I. Finding the MAXIMUM and MINIMUM values of a function in an interval.
MATH 6 fMin(expression, variable, lower, upper, tolerance)
This command returns the value of x at which the local minimum of the function
occurs back to in the given interval.
MATH 7 fMax(expression, variable, lower, upper, tolerance)
This command returns the value of x at which the local maximum of the function
occurs in the given interval.
Exercise 1. Find the maximum value of the function y   x 2  6 x  8 and show that the
maximum occurs at a critical value, that is, the derivative is either zero or undefined at a
critical point.
 Y= input Y1=-x^2+6x-8 2nd QUIT

VARS

Y1(fMAX(VARS Y-VARS 1
1 )

Y1(fMAX(Y1, X, -EE9, EE9) )
ENTER
Y-VARS
1
1 (to select Y1)
Y1( MATH
7)
answer: 1
Check the result by using 2nd CALC

WINDOW

2nd CALC 4

Left bound? ( Move to a point to the left of the maximum or enter a value of x for a
Xmin=-10
Xmax=10
Ymin=-20
Ymax=5 GRAPH
point to the left of the maximum, say 0 ) ENTER

Right bound? ( Move to a point to the right of the maximum or enter a value of x for
a point to the right of the maximum, say 5 ) ENTER

Guess? ( Move to a point near the maximum or enter a value of x for a point to near
the maximum, say 3 ) ENTER

answer: x= 3, y=1
MATH 8 nDeriv(Y1, X, 3) ENTER
answer: 0
The derivative at x=3 is zero, therefore the point (3, 1) is a critical point.
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Exercise 2. Find the minimum value of the function y  x 2  6 x  8 .

blue Y=

MATH 6 fMin( VARS Y-VARS

VARS
input
Y2=x^2+6x-8
Y-VARS
1
2
2nd QUIT
1 2, x, -EE9, EE9) answer x=-3
Y2( -3) =-17
Show that the minimum occurs at a critical value, that is, the derivative is either zero or
undefined.

MATH 8 nDeriv( VARS Y-VARs 1 2 , X, -3) ENTER ) answer: 0
The derivative at x=3 is zero, therefore the point (3, 1) is a critical point.
y  x 4  3x 2  2 x  3
Exercise 3. Consider the function
a) Find the point where the minimum value of the function occurs.

Y=

MATH

fMin(Y1, X, -EE8, EE8)

VARS Y-VARS 1
2nd QUIT
Y1=x^4 – 3x^2 +2x -3
6
fMin( VARS Y-VARS 1
1
ENTER
1
answer:-1.366024315
Y1( 2nd ANS )
ENTER
answer: -7.848076211
The point where the minimum occurs is (-1.366, -7.848)
b) Find the point between x=-1 and x= .75 where a maximum occurs.

MATH

fMax(Y1, X, -1, .75)

VARS Y-VARS 1 1 Y1( 2nd ANS )
7 fMax(VARS
Y-VARS 1
ENTER
1
answer: x=.3660258811
ENTER
answer: y=-2.651923789
The maximum between x=-1 and x=.75 occurs at (.366, -2.6520)
c) Find the values of x where the derivative is 0

MATH 0  CLEAR

MATH

 to A =-5 (any guess)

A=0 (2nd guess) ALPHA SOLVE

A=5 (3rd guess) ALPHA SOLVE answer: .99999933332
8 nDeriv( VARS Y-VARS 1
ALPHA SOLVE
1 ,
X , A)
answer: -1.366025070
answer: .36602573710
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ENTER But
x
Exercise 4. Consider the function y  xe
a) Draw the graph of the function, the derivative and the second derivative.
 Y=
Y1=xe^(-x) , Y2=MATH 8 nDeriv(VAR Y-VAR 1 1, x, x )

Y3=MATH 8 nDerv(VARS Y-VAR 1 2, x, x )

WINDOW :
Xmin=-3
Xmax=5
Ymin=-3 Ymax=3 GRAPH
2nd QUIT
b) Find the maximum value.

MATH

VARS Y-VARS 1 1 Y1( 2nd ANS ) ENTER
7
fMAX(Y1, x, -3, 5) ENTER
answer: x=1
answer: 0.3678794412
The maximum is 0.367879… and occurs at x=1
c) Show that the second derivative is negative at x = 1.

MATH 8 nDeriv( VARS
Y-VARS 1 3 ,x, 1) ENTER answer: -0.3678798075
d) Find the value(s) of x where the second derivative is zero.

MATH 0  CLEAR Eqn: 0 = VARS Y-VARS 1

x=1 (any guess) ALPHA SOLVE
3
ENTER
answer: x = 2
e) Use the second derivative test to show that there is a point of inflection at x=2

VARS Y-VARS 1 3 Y3(1) ENTER

2nd ENTRY (replace 1 by 3) Y3(3) ENTER answer:0.049787055, concave up.
answer: -0.367879075, concave down.
There is a change of concavity at x = 2, therefore we have a point of inflection at x = 2.
f) Find the point of inflection

VARS Y-VARS 1
1
Y1(2) ENTER
answer: 0.2706705665
The point of inflection is (2, 0.2706705665)
g) Find the average rate of change of the function between x=3 and x=7, that is, find
Y1 (7)  Y1 (3)
73

(VARS Y-VARS 1 1
Y1(7) - VARS Y-VARS 1 1 Y1(3))  (7-3) ENTER
The function is decreasing at an average rate of -.0357445078 units per unit change of X.
h) Find the instantaneous rate of change of the function at x-7.

VARS Y-VARS 1 2 Y2(7) ENTER answer: -.0054712924 units per unit
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Exercise 5. Consider the function y=x LN(x)
a) Graph the function

Y=
Y1=xLN(x) WINDOW: Xmin = -2 Xmax = 2 Ymin = -1

GRAPH 2nd QUIT
b) Use the first derivative to find the critical values of the function.

Y=

MATH 0  CLEAR

Let x=2 (or any guess) ALPHA SOLVE
Y2= MATH
8 Y2=nDeriv(VARS
Ymax = 2
Y-VARS 1 1, x, x) 2nd QUIT
VARS Y-VARS 1
2
ENTER
answer: x=0.3678798942 2nd QUIT
c) Find the critical point of the function.

VARS Y-VARS
1 1
Y1(x)
ENTER
answer: -0.3678794412
answer: the critical point is 0(.368, -0.368)
d) Test the critical point for maximum or minimum by using the first derivative test
(Use any value of x before the critical value)

VARS Y-VARS 1 2 Y2(0.25)
ENTER
answer: y  = -0.386297 < 0, the function is decreasing before the critical point

(Use any value of x after the critical value) 2nd ENTRY , (change x to 0.4)

Y2(0.4)
ENTER
answer: y  =0.083708>0, the function is increasing after the
critical point, the point is minimum point.
e) Use the second derivative test to verify that the point (0.368, -0.368) is a minimum

MATH 8
nDeriv(VARS Y-VARS 1 2, x, 0.368)
ENTER
answer: y  =0.083708>0, the function is concave up at the critical point, therefore the
critical point is a minimum point.
e) Show that lim xLN ( x)  0
x 0

VARS Y-VARS 1 1 Y1(0.01)= -0.0460517019

2nd ENTRY Y1(0.001)=-.0069077553

2nd ENTRY Y1(0.00001)=-.000115129255

2nd ENTRY Y1(10^-10)=-.00000000230258509
The limit is zero.
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x2
Exercise 6. Consider the function y  x
e
a) Graph the functionY=
Y1=x^2  e^(x)
 WINDOW: Xmin=-2 Xmax = 10 Ymin=-1 Ymax=4 GRAPH
b) Use the first derivative to find the critical values of the function.
 Y=
Y2= MATH 8 Y2=nDeriv(VAR Y-VARS 1 1, x, x) 2nd QUIT
 MATH 0  CLEAR VAR Y-VARS 1 2 ENTER Let x=2 (or any guess)
 ALPHA SOLVE
answer: x = 2
There is a critical value at x = 2. Let x=-2 (or any other guess)
 ALPHA SOLVE
answer: x = 4.99999…E-7  0. There is a critical point at x = 0
c) Find the critical points of the function.

VAR Y-VARS

2nd ENTRY (change x to 2) Y1(2) ENTER answer: a critical point is (2, .54)
1 1 Y1(0) ENTER
answer: a critical point is (0, 0)
d) Test the critical point for maximum or minimum using the second derivative test.Y=

Y3= MATH 8 nDeriv(VAR

2nd QUIT VAR Y-VARS 1 3
Y-VARS 1 2, x, x)
Y3(0) = 2 > 0. The function is concave up at x=0,
therefore there is a minimum at x = 0 2nd ENTRY (change x to 2) Y3(2) = -0.27<0.
The function is concave down at x = 2, therefore there is a maximum at x=2
e) Find the points of inflection

MATH 0  CLEAR VAR Y-VARS 1

ALPHA SOLVE
3 ENTER Let x = -2 (or any guess)
answer: x=0.585787
There is a possible point of inflection at x=0.585787
 Let x=2 (or any other guess) ALPHA SOLVE
answer: x = 3.4142139990
There is a possible point of inflection at x = 3.4142
 VAR Y-VARS 1 3 Y3(0) ENTER y  (0)=2>0 (concave up)

2nd ENTRY
Y3(2) ENTER
y  (2) = -0.27067<0 (concave down)
y  (2) =0.036631250>0 (concave up)
 2nd ENTRY Y3(4) ENTER
We have a point of inflections at x=0.585787 because the concavity changes from
positive to negative.
We have a point of inflections at x= 3.4142139990 because the concavity changes from
negative to positive.
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