www.MasterMathMentor.com Stu Schwartz AP Calculus – Functions Practice Test 1. Show that Rolle’s Theorem hold between x = 0 and x = 1 for fx () = x 3 " x + 5 . 2. Below is a graph of fx () . Place dots on the curve at the approximate locations that satisfy the mean-value theorem on [-4, 4]. 3. Find the value(s) of x that satisfy the mean-value theorem for fx () = 8 x " x 2 + 1 on [-1,3]. 4. To the right is a graph of " f x () . Determine what a graph of fx () might look like. Create sign charts to show your logic.
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www.MasterMathMentor.com Stu Schwartz
AP Calculus – Functions Practice Test 1. Show that Rolle’s Theorem hold between x = 0 and x = 1 for
!
f x( ) = x3" x + 5 .
2. Below is a graph of
!
f x( ) . Place dots on the curve at the approximate locations that satisfy the mean-value theorem on [-4, 4].
3. Find the value(s) of x that satisfy the mean-value theorem for
!
f x( ) = 8x " x 2 +1 on [-1,3].
4. To the right is a graph of
!
" f x( ). Determine what a graph of
!
f x( ) might look like. Create sign charts to show your logic.
www.MasterMathMentor.com Stu Schwartz
5. Below is a graph of
!
f x( ) shown on an interval. The graph of f has a horizontal tangent at c, d, f, and i. In the chart, place either a positive sign (+), negative sign (-) or zero (0) at the points a – j for
!
f x( ), " f x( ) and " " f x( ). If there is a relative minimum, relative maximum, absolute minimum, absolute maximum or possible inflection point on the interval at these points, put an x in the appropriate column.
Pt
!
f x( )
!
" f x( )
!
" " f x( ) Inflection pt.
Relative minimum
Relative maximum
Absolute Minimum
Absolute Maximum
a b c d e f g h i j
6. The figure to the right shows the graph of
!
" f , the derivative of the function f, for -6 ≤ x ≤ 6.
a) Find all values of x, for -6 < x < 6, at which f attains a relative maximum and relative minimum. Justify your answer. b) Find all values of x, for -6 < x < 6, at which f has an inflection point. Justify your answer.
www.MasterMathMentor.com Stu Schwartz
7. For the given function
!
f x( ) = 6x 2 " x 3 "1, find the x-values where
!
f x( ) attains a relative minimum, relative maximum, and inflection points, if any. Justify answers.
8. For the given function
!
f x( ) =x2 +1
x2"16
, find the intervals where the function is increasing and decreasing.
Justify your answer.
9. Find the absolute maximum and absolute minimum values of
!
f x( ) = x3 + 6x
2 +1 on "5,3[ ]. Be sure to state both what the relative extrema are and where they occur.
www.MasterMathMentor.com Stu Schwartz
AP Calculus – Functions Practice Test - Solutions 1. Show that Rolle’s Theorem hold between x = 0 and x = 1 for
!
f x( ) = x3" x + 5 .
!
f 0( ) = 5
f 1( ) =1"1+ 5 = 5
!
" f x( ) = 3x2#1= 0
3x2 =1
x = ±1
3 so x =
1
3
2. Below is a graph of
!
f x( ) . Place dots on the curve at the approximate locations that satisfy the mean-value theorem on [-4, 4].
3. Find the value(s) of x that satisfy the mean-value theorem for
!
f x( ) = 8x " x 2 +1 on [-1,3].
!
" f x( ) = 8 # 2x =f 3( ) # f #1( )
3+1=16 # #8( )
4
8 # 2x =24
4= 6
2 = 2x $ x =1
4. To the right is a graph of
!
" f x( ). Determine what a graph of
!
f x( ) might look like. Create sign charts to show your logic.
www.MasterMathMentor.com Stu Schwartz
5. Below is a graph of
!
f x( ) shown on an interval. The graph of f has a horizontal tangent at c, d, f, and i. In the chart, place either a positive sign (+), negative sign (-) or zero (0) at the points a – j for
!
f x( ), " f x( ) and " " f x( ). If there is a relative minimum, relative maximum, absolute minimum, absolute maximum or possible inflection point on the interval at these points, put an x in the appropriate column.
Pt
!
f x( )
!
" f x( )
!
" " f x( ) Inflection pt.
Relative minimum
Relative maximum
Absolute Minimum
Absolute Maximum
a - + - b 0 + - c + 0 - x d 0 0 + e + + 0 x f + 0 - x g 0 - - h - - 0 x i - 0 + x x j + + + x
6. The figure to the right shows the graph of
!
" f , the derivative of the function f, for -6 ≤ x ≤ 6.
a) Find all values of x, for -6 < x < 6, at which f attains a relative maximum and relative minimum. Justify your answer.
!
Relative maximum: x = 4 because
" f switches from positive to negative there.
Relative minimum: None because at no point
does " f switch from negative to positive.
b) Find all values of x, for -6 < x < 6, at which f has an inflection point. Justify your answer.
!
Inflection pts : x = "2,2,3,5 because # # f
switches sign at these values.
www.MasterMathMentor.com Stu Schwartz
7. For the given function
!
f x( ) = 6x 2 " x 3 "1, find the x-values where
!
f x( ) attains a relative minimum, relative maximum, and inflection points, if any. Justify answers.
!
" f x( ) =12x # 3x2 = 0
3x 4 # x( ) = 0
x = 0,x = 4
Rel min : x = 0 as " f switches from negative to positive
Rel max : x = 4 as " f switches from positive to negative
!
" " f x( ) =12 # 6x = 0
6 2 # x( ) = 0
x = 2
Inflection pt : x = 2 as " " f switches signs
8. For the given function
!
f x( ) =x2 +1
x2"16
, find the intervals where the function is increasing and decreasing.
Justify your answer.
!
" f x( ) =x
2#16( )2x # x
2 +1( )2x
x2#16( )
2=
2x3# 32x # 2x
2# 2x
x2#16( )
2=
#34x
x + 4( ) x # 4( )[ ]2
Critical values : x = 0, x = #4,x = 4
!
Function increasing on "#,"4( ), "4,0( ) as $ f > 0 on those intervals
Function decreasing on 0,#( ) as $ f < 0 on those intervals
9. Find the absolute maximum and absolute minimum values of
!
f x( ) = x3 + 6x
2 +1 on "5,3[ ]. Be sure to state both what the relative extrema are and where they occur.
!
" f x( ) = 3x2 +12x = 3x x + 4( )
Critical values : x = 0,x = #4
f 0( ) = 0 # 0 +1=1
f #4( ) = #4( )3
+ 6 #4( )2
+1= #64 + 6 16( ) +1= 33# 64 + 96 +1= 33
f #5( ) = #5( )3
+ 6 #5( )2
+1= #125 + 6 25( ) +1= #125 +150 +1= 26
f 3( ) = 3( )3
+ 6 3( )2
+1 = 27 + 6 9( ) +1= 27 + 54 +1= 82
Absolute minimum =1 at x = 0 Absolute maximum = 33 at x = #4
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