Practice Test
Group Members: Mina, Kunjan, Hiram, Zander
Chapter 2
1. Find the lim f(x) when f(x) = cos (4x + 4)
2.
a.
Find the vertical asymptotes
f(x)=tan(4x)sin(4x)
(3.2)
The graph of y=f(x) over the closed domain 2≤x≤8 is shown below. At what domain points does the
function appear to be…
a.
differentiable?
b.
c.
continuous but not differentiable?
neither continuous nor differentiable?
(3.3)
1. Find y1by applying the Product Rule, and then find y1by multiplying the factors to
produce a sum of simpler terms to differentiate
a.
y = (4x2+9)(9x-7+4x)
b.
find what y1=
2.
Find dydxif y= -8x7+1
(3.4)
The graph of f ‘(x) below shows the motion of a vampire hunter at night.
1. When is the vampire hunter…
a.
Moving forward?
b. Moving backward?
c. Speeding up?
d. Slowing Down?
e. At his greatest speed? In which direction?
f. Changing direction?
g. Moving at a constant speed?
2.
Use your pencil to draw a the equation of the speed of f(x) on the graph above.
(3.5)
1. Using the position function y = 12 sin(x) - 3 cos(x), find
a.
Velocity
b. Acceleration
c. Jerk
2.
3.
What is the equation of the line tangent to y = tan(x) + sec(x) at x = 2?
What is the equation of the normal?
(3.6)
1.
a.
2.
.
Use the given substitution to find ddx
y= cos(2x-9) , u=2x-9
Use the given substitution to find
y=sin(5x), u=5x
(3.7)
1. This is the equation (only in quadrant 1) of the Aphid that is infecting Mr. Dub’s tree:
2t2-s2(t)-7=0given in position, s (in inches), at any time, t (in seconds).
a.
Find the Aphid’s velocity equation
b.
Evaluate the Aphid’s velocity at t = 4
2.
Find d2ydx2if 2x3-3y2=8
(3.8)
Find the derivative of y=sec-1(2s3+4) with respect to s.
(3.9)
The spread of flu in a certain school t days after the flu first started to spread is modeled by the
equation P(t)=2001+e5-twhere P(t) is the total number of students infected t days after it begins
to spread.
a. Estimate the initial number of infected students
b. How fast is the flu spreading after 4 days?
c. When will the flu reach its maximum rate and what is it?
Answer Key
Chapter 2 Answers
1. Use a graph of a table of values for f(x)=cos (4x+ 4)
a.
x
f(x)
10
-0.9211
20
-0.9801
50
-0.9968
b. The table suggests that as x,the value of f(x) approaches 1
c.
Therefore, xf(x)=1
3.
Write f(x) in terms of sine and cosine,and determine where the denominator is equal to
zero. These places might correspond to the vertical asymptotes of y=f(x)
.
1cos(4x), the denominator will equal 0 whenever cos 4x=0
i.
when cos 4x=0 , x=8+ n4 , n any integer
a.
A check of the graph shows that the vertical asymptotes occur at x=8+ n4
(3.1)
3.2 Answers
a.
The function f is not differentiable at any point in its domain that is a corner, cusp,
vertical tangent, jump discontinuity or removable discontinuity. The function appears to have a
cusp at x=4, and discontinuities at x=5 and x=6. Therefore, the graph is not differentiable at these
3 points.
b.
To determine at what point(s) the domain of the function is continuous but not
differentiable, consider the previously determined non-differentiable points x=4, x=5, and x=6.
Because x=5 and x=6 are discontinuities, they are not continuous, and therefore only x=4 which
is cusp is continuous but not differentiable.
c.
To determine at which point(s) in the domain the function appears to be neither
continuous nor differentiable, once more consider the previously determined non-differentiable
points x=4, x=5, and x=6. At points x=5 and x=6 the graph has a break and therefore these points
are neither continuous nor differentiable.
3.3 Answers
1. First, find y1 by applying the Derivative Product Rule, which states that if u and v are
differentiable at x, then so is their product uv, and ddx(uv)=udvdx+vdudx
2. y1=ddx((4x2+9)(9x-7+4x)
3. =(4x2+9)ddx(9x-7+4x)+ (9x-7+4x)ddx(4x2+9)
a.
to find ddx(9x-7+4x), use the Derivative Sum Rule,which states if u and v are
differentiable functions of x,then their sum u+v is differentiable at every point where u and v are
both differentiable. At such points, ddx(u+v) = dudx+ dvdx
4. ddx(9x-7+4x)=ddx(9x) + ddx(-7) + ddx(4x)
.
to find ddx(9x), use the Constant Multiple Rule, which states if u is a differentiable
function of x, and ci is a constant, then ddx(cu) = cdudx
a.
ddx(9x)=9ddx(x)=9
5. To find ddx(-7), use the derivative of a Constant Function Rule,which states that if f has
the constant value f(x)=c,then dfdx= ddx*c =0
.
ddx(-7)=0
a.
before evaluating ddx(4x),write 4x using exponents
i.
4x=4x-1
ii.
to find ddx(4x-1), use the Constant Multiple Rule and the Power Rule for Negative
Integers. The Power Rule for Negative Integers states that if n is a negative integer and x 0, then
ddx(xn) = nxn-1
iii.
ddx(4x-1) = -4x2
iv.
Thus, ddx(9x-7+4x) = 9-4x2
6. To find , ddx(4x2+ 9 ), use the Derivative Sum Rule
.
, ddx(4x2+ 9 ) = ddx(4x2) + ddx(9)
a.
To find ddx(4x2), use the Constant Multiple Rule and the Power Rule for Positive
Integers. The Power Rule for Positive Integers states that if n is a positive integer, then ddx
(xn)=nxn-1
.
ddx(4x2) = 8x
7. To find ddx(9), use the Derivative of a Constant Function Rule
.
ddx(9)= 0
a.
Thus, ddx(4x2+9) = 8x
8. Now simplify
.
(4x2+9) ddx(9x-7+4x) ddx (4x2+9)
a.
=(4x2+9)(9-4x2) + (9x-7 + 4x) (8x) = 108x2-56x+97-36x2
b.
This,by applying the Derivative Product Rule, y1=108x2-56x+97-36x2
9. Now find y1by multiplying the factors to produce a sum of simpler terms to differentiate.
Multiply
.
(4x2+9)(9x-7 + 4x) = 36x3-28x2+97-63+36x
10. To find ddx (36x3-28x2+97-63+36x), use the Derivative Sum Rule
.
=ddx(36x3)ddx(-28x2)+ddx (97)+ ddx(-63)+ddx(36x)
11. Find the derivatives of each of the differentiable function
.
ddx(36x3)=108x2Constant Multiple Rule, Power Rule
a.
ddx(-28x2)=-56xConstant Multiple Rule, Power Rule
b.
c.
d.
.
ddx(97x) = 97 Constant Multiple Rule, Power Rule
ddx(-63) = 0 Derivative of a Constant Function Rule
ddx(36x) = - 36x2
12. Thus, by multiplying the factors to produce a sum of simpler terms to differentiate
y1=108x2-56x+97-36x2
1. To find dydx, first use the derivative sum rule,which states if u and v are differentiable
functions of x,then their sum u+v is differentiable at every point where u and v are both
differentiable. At such points, dydx(u+v)=dudx+ dvdx
2. dydx=ddx(-8x7+1) + ddx(1)
3. To find ddx(-8x7)+ddx(1)
4. To find ddx(-8x7), use the constant multiple rule and the power rule for positive integer
powers of x
5. The constant multiple rules says that if u is a differentiable function of x, and c is a
constant,then ddx(cu)=cdudx
6. The power rule for positive integers states that if n is a positive integer, then
ddx(xn)=nxn-1
a.
ddx(-8x7)=-8ddx(x7)= 8*7x6=-56x6
7. To find ddx(1), use the rule for the derivative of a constant function, which states that if f
has the constant value f(x)=c,then dfdx=ddx*c = 0
.
ddx(1)=0
a.
Thus, dydx= ddx(-8x7+1) = -56x2
3.4 Answers
1. When is the vampire hunter…
h. Moving forward?
(1,6)u(8,9)
j. Speeding up?
(1,2)u(3,5)u(6,7)u(8,9)
l. When is he at his greatest speed? In which direction?
5 and 9 forward. 7 backward.
n. Moving at a constant speed?
Never
2.
3.5 Answers
1. Using the position function y = 12 sin(x) - 3 cos(x), find
i. Moving backward?
(0,1)u(6,8)
k. Slowing Down?
(0,1)u(2,3)u(5,6)u(7,8)
m. Changing direction?
@x = 1, 6 and 8
a.
Velocity
12 cos(x) + 3 sin(x)
b. Acceleration
-12 sin(x) + 3 cos(x)
c. Jerk
-12 cos(x) - 3 sin(x)
2.
What is the equation of the line tangent to y = tan(x) + sec(x) at x = 2?
y = tan(2) + sec(2)= 01 +11 = 1
Now we know our point is at (,1)
y' = sec2(x) + sec(x)tan(x)
= (1/1)2 + 1101
= 1+0 = 1
Derivative of original equation
Trigonometric ratios
Simplify
In point-slope form our equation would be:
y - 1 = 1(x-)
3.
Normal
y - 1 = -1(x-)
Negative Reciprocal
(3.6)
1. To find dydx, use the chain rule,which states if f(u) is differentiable at the point u=g(x)
and g(x) is differentiable at x, then the composite function (f o g)(x) = f(g(x)) is
differentiable at x, and (f o g)'(x) = f '(g(x)) * g ' * (x)
2. Use Leibniz’s notation: if y-f(u) and u=g(x), then dydx=dydu*dudx,where dydu is
evaluated at u=g(x)
a.
= ddu(cos u) * ddx(2x-9)
b.
Use the definitions of the derivatives of the trigonometric function to find ddu(cos u) = sin u
3. Find the derivative of (2x-9)
.
ddx(2x-9)=2
4. Substitute these derivatives into dydxand simplify
.
i.
ii.
dydx= ddu(cos u) * ddx(2x-9)
= -sin u * (2)
= -2sin (2x-9)
5. Thus, dydx= -2 sin (2x-9)
1. To find dydx, use the chain rule, which states if f(u) is differentiable at the point u=g(x)
and g(x) is differentiable at x,then composite function (f o g) (x) =f(g(x)) is differentiable
at x, and (f o g)'(x) = f '(g(x)) * g ' * (x). In Leibniz’s notation: if y-f(u) and u=g(x), then
dydx=dydu*dudx,where dydu is evaluated at u=g(x)
2. Substitute function into the chain rule
a.
dydx=dydu*dudx= ddu(sin u) * ddx(5*x)
3. Use the definition of the derivatives of the trigonometric functions to find ddu(sin u)
.
ddu(sin u) = cos u
4. Find the derivative of (5x)
.
ddx= ddu(sin u) * ddx(5x)
a.
= cos u * (5)
b.
= 5 cos (5)
5. Therefore, dydx= (5)cos (5x)
(3.7)
a.
Find the Aphid’s velocity equation
Use implicit differentiation
ddx2t2-ddxs2(t)-ddx(7)=ddx(0)
4t-(2s(t)s'(t))-0=0
-(2s(t)s'(t))=-4t
s'(t)=-4t-2s(t)
s'(t)=4t2s(t)
s'(t)=2ts(t)
Solve the original equation for s(t)
2t2-s2(t)-7=0
s(t)=(-7+2t2)
Plug in s(t) into the velocity equation
s'(t)=2t(-7+2t2)
The Aphid’s velocity equation is 2t(-7+2t2)
b. Evaluate the Aphid’s velocity at t = 4
s'(t)=2t(-7+2t2)
s'(4)=24(-7+242)
s'(4)=8(-7+216)
s'(4)=8(-7+32)
s'(4)=825)
s'(4)=85
At t = 4, the velocity is 85in./sec
Find d2ydx2if 2x3-3y2=8
Use implicit differentiation to first find dydx
dydx2x3- dydx3y2= dydx8
6x2-6yy'=0
x2-yy'=0
yy'=(-x2)
y'=(-x2)(-y)
y'=x2y
Now, use the quotient rule to find d2ydx2
y''= dydx(x2y)
Quotient rule: a'b-ab'b2;
a=x2,
a'=2x, b'=y'
y''= 2xy-x2y'y2
y''= 2xy-x2y2y'
b=y
Substitute y'=x2y to express y''in terms of xand y
y''= 2xy-x2y2y'
y''= 2xy-x2y2x2y
y''= 2xy-x4y3
(3.8)
For u, a differentiable function of s, the derivative of sec-1u is 1uu2-1duds.
The function u(s) is 2s3+4.
duds= 6s2
substituting the derivative of sec-1(2s3+4) is 6s22s3+4(2s3+4)2-1, or 6s22s3+44s6+16s3+15
(3.9)
a.
set t=0
2001+e5-0=1.3, round down to 1
b. Finding the derivative to find the rate:
(-200(-e5-t))(1+e5-t)2
Plug in 4 for t to get 39 students/day