Name: _________________________________________
Section (assigned): ______ Section (attended): _______
Math 10A Midterm 1
January 30, 2009
Notes allowed (one page), graphing calculators allowed, show all work
50 points total, each problem worth 10 points
1. Find a quadratic function, f ( x ) = ax 2 + bx + c , such that the point (1, 0 ) is the
vertex of the graph of f ( x ) and the y-intercept of the graph is ( 0, 2 ) . (Hint: the
vertex form of the quadratic equation is y = a ( x ! h ) + k , where the vertex is the
point ( h, k ) .): (Precalculus Review, 1.6.22)
2
(5 points) Substitute vertex and y-intercept into vertex form of quadratic equation
and solve for a, i.e., 2 = a ( 0 ! 1) + 0 and a = 2 .
(3 points) Substitute a and vertex into vertex form of quadratic equation, i.e.,
2
y = 2 ( x ! 1) + 0 .
(2 points) Convert vertex form to general form of quadratic equation, i.e.,
2
y = 2x 2 ! 4x + 2 .
2. If possible, choose constants a and b so that the following function is continuous
and differentiable at x = 0 : (Lecture 1/23)
x<0
"$ x + a
g( x) = # 2
%$ x + bx! !!!!!x ! 0
(5 points) Determine a and/or b such that g is continuous at x = 0 . Substitute
x = 0 into both "pieces" and set them equal to each other, namely
0 + a = 0 2 + b ! 0 . Solve for a ( a = 0 ).
(5 points) Determine a and/or b such that g is continuous at x = 0 . Find
derivatives of each piece (using rules is OK), evaluate each derivative at
x = 0 , and set them equal to each other, namely 1 = 2 ( 0 ) + b . Solve for b
( b = 1 ).
3. Given the function g ( x ) = !2x 2 + x ! 3 : (2.3.15)
a. Estimate g! ( x ) using the average rate of change formula. Let h = 0.1.
(4 points)
2
2
$
&
g ( x + .1) # g ( x ) % #2 ( x + .1) + ( x + .1) # 3' # $% #2x + x # 3&'
g! ( x ) "
=
=
.1
.1
(
)
" !2 x 2 + .2x + .01 + x + .1 ! 3$ ! " !2x 2 + x ! 3$
%
#
% #
=
.1
" !2x 2 ! .4x ! .02 + x + .1 ! 3$ ! " !2x 2 + x ! 3$
#
% #
%=
.1
!.4x ! .02 + .1
= !4x ! .2 + 1 = !4x + 0.8 .
.1
g ( x + h) # g ( x)
.
h"0
h
b. Calculate g! ( x ) using the limit definition, g! ( x ) = lim
DO NOT USE THE POWER RULE.
(4 points)
$ #2 ( x + h )2 + ( x + h ) # 3& # $ #2x 2 + x # 3&
g ( x + h) # g ( x)
'
' %
g! ( x ) = lim
= lim %
=
h"0
h"0
h
h
(
)
# "2 x 2 + 2hx + h 2 + x + h " 3% " # "2x 2 + x " 3%
&
& $
lim $
=
h!0
h
# "2x 2 " 4hx " 2h 2 + x + h " 3% " # "2x 2 + x " 3%
& $
&=
lim $
h!0
h
"4hx " h 2 + h
= lim " 4x " h + 1 = "4x + 1 .
h!0
h!0
h
lim
c. Calculate g! ( "2 ) .
(2 points) g! ( "2 ) = "4 ( "2 ) + 1 = 9 .
4. Let f ( t ) be the number of centimeters of rainfall that has fallen since midnight,
where t is the time in hours. Eight centimeters of rain have fallen by 10 AM.
(2.4.15)
a. Interpret in practical terms, giving units: f ! (10 ) = 2 .
(5 points) The rate that rain is falling at 10 AM is 2 cm/hr.
b. Determine the equation of the tangent line to the graph of f ( t ) at t = 10 .
(5 points) Use the rate from part (a) and the "point", (10, 8 ) , given in the
problem statement to find the equation of the tangent line at t = 10 , e.g.,
y = 2(t ! 10) + 8 = 2t ! 12 .
5. The figure below is the graph of f ! ( x ) , the derivative of a function f ( x ) . THIS
IS NOT THE GRAPH OF f ( x ) . At which of the marked values of x is: (2.3.42,
2.5.21)
a. f ( x ) increasing
(4 points) f is increasing where f ! ( x ) is positive, namely at x3 , x4 , x5
b.
f !( x ) increasing
(3 points) The graph of f ! ( x ) is increasing at x2 , x3
c.
f !!( x ) < 0
(3 points) f !!( x ) < 0 when f ! ( x ) is decreasing, namely at x1, x4 , x5 .