[mean value theorem][section 13]

(1.)  f(x) = 1 + 2x2  ;  (-1,1)

f'(x) = 4x              take the derivative

f(-1) = 1 + 2(-1)2    f(1) = 1 + 2(1)2

[replace x with -1 and 1 in f]

f(-1) = 3          f(1) = 3          multiply, add

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = (3 - 3)/(1 - -1)    make substitutions

m = 0           subtract, add, divide

4x = 0      set 4x equal to 0
___  ___
4     4         divide each side by 4

x = 0         divide and cancel

result:  c = 0

(2.)  f(x) = 1/x ;   (1,4)         here is the problem

f(1) = 1/1  ;   f(4) = 1/4     replace x with 1 and 4

f(1) = 1  ;   f(4) = 1/4        divide

m = (y2 - y1)/(x2 - x1) use the slope formula

m = [(1/4) - 1]/(4 - 1)    make substitutions

m = (-3/4)/(3)            subtract

m = -1/4                   divide

f'(x) = -1/x2        take the derivative of x

-1/x2 = -1/4           set the derivative equal to -1/4

1/x2 = 1/4             multiply each side by -1

x2 = 4                   take reciprocals

x = 2           take square roots

result:  c = 2
_
(3.)  f(x) = √x  ;  (1,4)           here is the problem
_               _
f(1) =
1   ;   f(4) = 4     replace x with 1 & 4

f(1) = 1   ;   f(4) = 2    take square roots

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = (2 - 1)/(4 - 1)   make substitutions

m = 1/3                  subtract

f'(x) = (1/2)x-1/2      take the derivative of f

1/3 = (1/2)x-1/2        replace f'(x) with 1/3

2 = 3x-1/2      multiply each side by 6, cancel
___ ______
3     3              divide each side by 3

2/3 = x-1/2             cancel

3/2 = x1/2     take the reciprocal of each side

9/4 = x         square each side

result:  c = 9/4
_
(4.)  f(x) = \3/x  ;  (-8,-1)           here is the problem
__                __
f(-8) = \3/-8  ;   f(-1) = \3/-1

[replace x with -8][replace x with -1]

f(-8) = -2          f(-1) = -1     take cube roots

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = (-1 - -2)/(-1 - -8)   make substitutions

f'(x) = (1/3)x-2/3     take the derivative of f

1/7 =(1/3)x-2/3           replace f' with 1/7

3 = 7x-2/3           multiply each side by 21, cancel
__ ______
7    7                 divide each side by 7

3/7 = x-2/3               cancel

7/3 = x2/3             take reciprocals of each side

x2/3 = 7/3        just rearrange like this

x2 = (7/3)3      cube each side
______             ______
x =   √(7/3)3        x = -√(7/3)3    take the sq rt of ea side
_______
result:  c = -√(7/3)3

(5.)  f(x) = x3  ;  (1,2)        here is the problem

f(1) = 13      f(2) = 23     replace x with 1 and with 2

f(1) = 1     f(2) = 8     cube 1 and cube 2

m = (y2 - y1)/(x2 - x1)   use the slope formula

m = (8 - 1)/(2 - 1)    make substitutions

m = 7                subtract, divide

f'(x) = 3x2          take the derivative of f

7 = 3x2            replace f' with 7
__  ___
3    3                divide each side by 3

7/3 = x2                cancel
___
x =
7/3               take square roots
____
result:  c =
7/3

(6.)  f(x) = x3 ;  (-2,2)

(7.)  f(x) = (x + 3)(x - 1)(x - 5) ;  (-3,1)

f(x) = (x + 3)(x2 - 6x + 5)  foil multiply combine like terms

f(x) = x3 - 6x2 + 5x + 3x2 - 18x + 15   multiply

f(x) = x3 - 3x2 - 13x + 15   combine like terms

f(-3) = (-3 + 3)(-3 - 1)(-3 - 5)   f(1) = (1+3)(1-1)(1-5)

[replace x with -3][replace x with 1]

f(-3) = 0   f(1) = 0     simplify

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = (0 - 0)/(1 - -3)   make substitutions

m = 0              subtract add divide

f'(x) = 3x2 - 6x - 13   take the derivative of f

3x2 - 6x - 13 = 0   set the derivative equal to 0

+ 16  +16    add 16 to each side
____________________
3x2 - 6x + 3 = 16          add

3(x2 - 2x + 1) = 16             factor

3(x - 1)2 = 16                 factor

(x - 1)2 = 16/3        divide each side by 3, cancel
_                 _
x - 1 = 4/
3     x - 1 = -4/3      take square roots

+ 1  +1            +1  +1     add 1 to each side
____________   ______________
_                    _
x =  1 + (4/
3)      x =  1 - (4/3)
_
result:  c = 1 - (4/
3)

(8.)  f(x) = x3 - 3x2 + 1 ;   (0,3)

f(0) = 03 - 3(0)2 + 1    f(3) = 33 - 3(3)2 + 1

[replace x with 0][replace x with 3]

f(0) = 1     f(3) = 1     simplify

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = (1 - 1)/(3 - 0)  make substitutions

m = 0               subtract and divide

f'(x) = 3x2 - 6x          take the derivative of f

3x2 - 6x = 0          set the derivative equal to 0

3x(x - 2) = 0             factor

x - 2 = 0    set this factor equal to 0

+  2  +2   add 2 to each side
___________

result: c = 2

(9.)  f(x) = cos x ; (0, pi/2)

f(0) = cos 0     f(pi/2) = cos (pi/2)

[replace x with 0][replace x with pi/2]

f(0) = 1           f(pi/2) = 0    use the unit circle

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = (0 - 1)/[(pi/2) - 0]      make substitutions

m = -2/pi          subtract, and take the reciprocal of

the bottom

f'(x) = -sin x        take the derivative of f

-sin x = -2/pi          set the derivative equal to -2/pi

sin x = 2/pi          multiply each side by -1

x = arcsin (2/pi)    take the arcsin of each side

result:   c = arcsin (2/pi)

(10.)  f(x) = sin 2x  ;   (0, pi/12)

f(0) = sin 2(0)    f(pi/12) = sin 2(pi/12)

[replace x with 0][replace x with pi/12]

f(0) = sin 0        f(pi/12) = sin (pi/6)  multiply, reduce

f(0) = 0      f(pi/12) = 1/2   use the unit circle

m = (y2 - y1)/(x2 - x1)  use the slope formula

m = [(1/2) - 0]/[(pi/12) - 0]   make substitutions

m = (1/2)/(pi/12)          subtract

m = (1/2)*(12/pi)     multiply by the reciprocal of the bottom

m = 6/pi               multiply and reduce

f'(x) = 2 cos 2x          take the derivative of f

2 cos 2x = 6/pi           set the derivative equal to 6/pi

cos 2x = (1/2)(6/pi)      multiply each side by 1/2, cancel

cos 2x = 3/pi                    divide

2x =  arccos (3/pi)         take the arccos of each side

x = (1/2)arccos (3/pi)    multiply each side by 1/2, cancel

result:  c = (1/2)arccos (3/pi)