AB Calculus Solutions to Sample 1
Detailed solutions and explanations to the AB calculus questions solutions in sample 1.

=
Solution
As h > 0
lim (e^{4} e^{h}  e^{4}) / h = (e^{4} e^{0}  e^{4}) / 0
= 0 / 0 , indeterminate form
Another approach is needed. Let f(x) = e^{x}. The given limit may be written as follows: as x > h
lim (e^{4} e^{h}  e^{4}) / h = lim (e^{4 + h}  e^{4}) / h = limit [ f(4+h)  f(4) ] / h
which is the definition of the first derivative of f(x) = e^{x} at x = 4. Hence as x > h
lim (e^{4} e^{h}  e^{4}) / h = e^{ 4}

The graph of function g defined by
will have vertical asymptotes at
Solution
Let us first simplify, if possible, the given rational function
g(x) = (x^{3} + 2x^{2}  3x) / (x^{2} + 2x  3)
= x (x^{2} + 2x  3) / (x^{2} + 2x  3) = x
Function g has no vertical asymptotes

Given that
find
Solution
Using the theorem that states that the limit of a sum is equal to the sum of the limits. Hence as x > 0,
lim (x + 4x^{2} + sin x) / 3x = lim (x / 3x) + lim (4x^{2} / 3x) + (1/3) sin x / x
Simplify
= lim (1/3) + lim (4x) + (1/3) sin x / x
= 1/3 + 0 + (1/3)*1 = 2/3

Function f is defined by
.
Find df(x) / dx.
Solution
Using the theorem that states that the derivative of a sum of functions is the sum of the derivatives, we can write
d/dx [ 2x^{3}sin(x) + (1/x)tan(x) + x sec(x) + 2 ]
= d/dx [ 2x^{3}sin(x) ] + d/dx [ (1/x)tan(x) ] + d/dx [ x sec(x) ] + d/dx [ 2 ]
we now calculate the derivative of each term above
d/dx [ 2x^{3}sin(x) ] = 2 [ 3x^{2} sin(x) + x^{3} cos(x) ] = 6x^{2} sin(x) + 2x^{3} cos(x)
d/dx [ (1/x)tan(x) ] =  (1/x^{2}) tan(x) + (1/x) sec^{2}(x))
d/dx [ x sec(x) ] = sec(x) + x sin(x) sec^{2}(x)
d/dx [ 2 ] = 0
Hence
df/dx = 6x^{2} sin(x) + 2x^{3} cos(x)  (1/x^{2})tan(x) + (1/x sec^{2}(x)) + sec(x) + x sin(x) sec^{2}(x)

Curve C is described by the equation 0.25 x^{2} + y^{2} = 9. Determine the y coordinates of the points on curve C whose tangent lines have slope equal to 1.
Solution
Let us calculate the first derivative. Differentiate both sides of the given equation
0.25 (2x) + 2 y y ' = 0
y ' =  0.5 x / (2 y)
We now solve the given equation 0.25 x^{2} + y^{2} = 9 for x
x = + or  sqrt [ (9  y^{2}) / 0.25 ]
Substitute x in y ' =  0.5 x / (2 y) by + or  sqrt [ (9  y^{2}) / 0.25 ]
y ' =  0.5 (+ or  sqrt [ (9  y^{2}) / 0.25 ]
/ (2 y)
The slope of the tangent is equal to 1. Hence
 0.5 (+ or  sqrt [ (9  y^{2}) / 0.25 ] )/ (2 y) = 1
Solve the above for y. Two solutions
y = 3 sqrt(5) / 5 , y =  3 sqrt(5) / 5
The graph of 0.25 x^{2} + y^{2} = 9 is shown below with the two tangent lines.

Find the solution to the differential equation dy/dx = cos(x) / y^{2} , where y(π/2) = 0.
Solution
The variable in the given differential equation may be separated as follows
y^{2} dy = cos(x) dx
Integrate both sides
∫ y^{2} dy = ∫ cos(x) dx
(1/3) y^{3} = sin(x) + C , constant of integration
We now use the condition y(π/2) = 0 to find the constant C
(1/3) y^{3}(π/2) = sin(π/2) + C
0 = 1 + C
C =  1
Substitute C by 1 in (1/3) y^{3} = sin(x) + C and solve for y
(1/3) y^{3} = sin(x)  1
y^{3} = 3(sin(x)  1)
y = (3 sin(x)  3)^{1/3}

Solution
Let u = cos x and therefore du/dx =  sin x. We now substitute cos x by u and sin x by du/dx in the given integral. Hence
∫cos^{4}(x)sin(x) dx = ∫ u^{4} (du/dx) dx
=  ∫ u^{4}du
= (1/5)u^{5} + C , C constant of integration
= (1/5)cos^{5}(x) + C

Solution
Let u = 2x and therefore du/dx = 2 or dx = du / 2. Hence the given integral becomes
d/dx ∫ _{3}^{2x}sin(t^{2} + 1) dt = 2 d/du ∫ _{3}^{u}sin(t^{2} + 1) dt
using the fundamental theorem of calculus, we obtain
= 2 sin(u^{2} + 1)
Substitute u by 2x
= 2 sin(4x^{2} + 1)

Solution
We first to analyze the signs of the expressions 4  x and 2  2x between the limits of integration 0 and 10. 4  x changes sign at x = 4 and 2  2x changes sign at x = 1.
for x between 0 and 4: 4  x is positive and hence 4  x = 4  x
for x between 4 and 10: 4  x is negative and hence 4  x = (4  x)
for x between 0 and 1: 2  2x is positive and hence 2  2x = 2  2x
for x between 1 and 10: 2  2x is negative and hence 2  2x = (2  2x)
We now rewrite the given integral as a sum of two integrals as follws.
∫ _{0}^{10} (4  x+2  2x) dx =
∫ _{0}^{10} (4  x) dx + ∫ _{0}^{10} (2  2x) dx
We now calculate each of the individual integrals above as follows.
∫ _{0}^{10} (4  x) dx =
∫ _{0}^{4} (4  x) dx + ∫ _{4}^{10} (4  x) dx = 8 + 18 = 26
and
∫ _{0}^{10} (2  2x) dx =
∫ _{0}^{1} (2  2x) dx + ∫ _{1}^{10} (2  2x) dx = 1 + 81 = 82
We now have
∫ _{0}^{10} (4  x) dx + ∫ _{0}^{10} (2  2x) dx = 26 + 82 = 108

Evaluate the integral
Solution
Let u = 5 + x^{3/4} and therefore du/dx = (3/4) 1/x^{1/4} and substitute in the given integral
∫ (5 + x^{3/4})^{9} / (x^{1/4}) dx = ∫
[ ( u^{9} ) / (x^{1/4}) ] (4/3) x^{1/4} du
= (4/3) ∫ u^{9} du
= (4/3) (1/10) u^{10}
= (2/15) (5 + x^{3/4})^{10}

Given that function h is defined by
find h'(x).
Solution
Let u = arctan(x^{3} + 1) + 2x. Hence function h can be written as
h(x) = u^{4}
h '(x) = 4 u^{3} u'
We now let v = arctan(x^{3} + 1) and calculate u '
u ' = (v ')( 1 / (1 + v^{2}) )
= (3x^{2}) / (1 + (x^{3} + 1)^{2})
= (3x^{2}) / (x^{6} + 2x^{3} + 2)
Hence
h '(x) = 4 (arctan(x^{3} + 1) + 2x)^{3} (3x^{2}) / (x^{6} + 2x^{3} + 2)

The graph of function h is shown below. How many zeros does the first derivative h' of h have?
Solution
Whenever the graph of h has a local maximum or local minimum h '(x) is equal to 0. The given graph has 3 local minima and 2 local maxima and therefore h ' has 5 zeros.

The graph of a polynomial f is shown below. If f' is the first derivative of f, then the remainder of the division of f'(x) by x  b is more likely to be equal to
Solution
The graph of f has a local maximum at b and therefore f'(b) = 0. Since f is a polynomial then f ' is also a polynomial function such that f '(b) = 0 and according to the remainder theorem the division of f '(x) by x  b is equal to o.

The set of all points (ln(t  2) , 3t), where t is a real number greater than 2, is the graph of
Solution
The given parametric equations may written as
x(t) = ln(t  2) and y(t) = 3t
Solve y(t) = 3t for t
t = y / 3
Substitute t by y / 3 in x(t) = ln(t  2)
x = ln(y / 3  2)
Solve for y
y/3  2 = e^{x}
y = 3 ( e^{x} + 2 )

Let P(x) = 2 x^{3} + K x + 1. Find K if the remainder of the division of P(x) by x  2 is equal to 10.
Solution
The Remainder theorem states that the division of P(x) by x  2 is equal to P(2). Hence
P(2) = 2 (2)^{3} + K (2) + 1 = 10
Solve for K
K =  7/2

Function f is defined by
.
where C is a constant. What must the value of C be equal to for function f to be continuous at x = 0?
Solution
For function f to be continuous at x = 0, lim f(x) as x approaches must be equal to f(0). We first find the limit of f(x) as x approaches 0. As x > 0,
lim [ sqrt(4x + 4)  sqrt(2x + 4) ] / 2x = 0 / 0 , indeterminate
Another approach is needed. Multiply numeartor and denominator by sqrt(4x + 4)  sqrt(2x + 4), simplify and find the limit. As x > 0,
lim [ sqrt(4x + 4)  sqrt(2x + 4) ] / 2x
= lim [ sqrt(4x + 4)  sqrt(2x + 4) ] [ sqrt(4x + 4) + sqrt(2x + 4) ]/ [ 2x [ sqrt(4x + 4) + sqrt(2x + 4) ] ]
= lim [4x + 4  2x  4] / [ 2x [ sqrt(4x + 4) + sqrt(2x + 4) ] ]
= lim 2x / [ 2x [ sqrt(4x + 4) + sqrt(2x + 4) ] ]
= lim 1 / [ sqrt(4x + 4) + sqrt(2x + 4) ]
= 1 / [ 2 + 2 ] = 1/4
In order for f to be continuous, we need to have
C = 1/4

f and g are functions such that f '(x) = g(x) and g '(x) = f(x). The second derivative of (f^{ . }g)(x) is equal to
A) f "(x) g "(x)
B) g '(x) g(x) + f(x) f '(x)
C) 4 g(x) f(x)
D) 2 g(x) f(x)
E) g(x) f(x)
Solution
First derivative of (f^{ . }g)(x)
(f^{ . }g) ' = f ' g + f g '
Second derivative of (f^{ . }g)(x)
(f^{ . }g) " = (f ' g + f g ') '
= f "g + f 'g' + f 'g' + f g" (I)
Note that since f '(x) = g(x) and g '(x) = f(x), we have
f " = g' and g " = f'
Substitute f " and g " in (I) above to obtain
(f^{ . }g) " = g 'g + f 'g' + f 'g' + f f'
We now substitute g ' by f and f ' by g to obtain
(f^{ . }g) " = fg + gf+ fg + f g = 4 fg

The average rate of change of the function f defined by f(x) = sin(x) + x on the closed interval [0 , pi] is equal to
Solution
The average rate of change of a function from a to b is defined by
(f(b)  f(a)) / (b  a)
Apply the above definition to the question above
(f(pi)  f(0)) / (pi  0) = [ (sin(pi) + pi)  (sin(0)+ 0) ] / (pi  0) = 1

The figure shows the graphs of y = sin(x) over half a period and the line y = 1/2. Find area of the shaded region.
.
Solution
The x coordinates of the points of intersection the line y = 1/2 and the graph of f are found by solving
sin(x) = 1/2 , for 0 ≤ x ≤ pi
solutions: x = pi / 6 and x = 5pi / 6
We first split the area to be calculated into 3 parts as shown below.
.
Calculate the area A adding the three parts (NOTING that the area in the middle is that of a rectangle):
A = ∫_{0}^{pi/3} sin(x) dx+ 1/2(5Pi/6  Pi/6) + ∫_{5pi/6}^{pi} sin(x) dx
= [  cos(x) ]_{0}^{pi/3} + pi / 3 + [ cos(x)]_{5pi/6}^{pi}
= 2 + pi / 3  sqrt(3)

Functions f, g and h are defined as follow: g(x) = f(x^{2}), f(x) = h(x^{3} + 1) and h'(x) = 2x + 1. g'(x) =
Solution
Let us express g(x) in terms of h
sin(x) = 1/2 , for 0 ≤ x ≤ pi
g(x) = f(x^{2}) = h( (x^{2})^{3} + 1 )
= h(x^{6} + 1)
Let u = x^{6} + 1. Hence
g(x) = h(u) , with u = x^{6} + 1
Use chain rule to write
g '(x) = (du/dx) (dh/du)
If h'(x) = 2x + 1, then
dh/du = h '(u) = 2u + 1
Hence
g '(x) = 6x^{5} [ 2u + 1 ]
Substitute u by x^{6} + 1 in g '(x)
g '(x) = 6x^{5} [ 2(x^{6} + 1) + 1 ] =
= 12 x^{11} + 18 x^{5}