Mr. Simonds’ MTH 251 Class
1.
2.
a.
g ′′ is positive over ( −2, 0 ) .
b.
g is concave down over ( −∞, − 2 ) and ( 0,1) .
c.
g is nondifferentiable at 3 .
d.
g ′ is nondifferentiable at −5 , −4 , −2 , 1 , and 3 .
e.
g is linear on (1,3) and ( 3, ∞ ) .
f.
g ′′ is increasing on ( −5, − 4 ) .
g.
g is increasing on ( −∞, ∞ ) .
h.
g ′′ is constant on ( −4, − 2 ) , (1,3) and ( 3, ∞ ) .
i.
g has no minimum value
j.
g ′′ ( x ) = 0 at 0 and over (1,3) and ( 3, ∞ ) .
k.
There’s no way of knowing where antiderivatives of g are increasing.
l.
Antiderivatives of
a.
c.
1
y ( t ) = − t −8/3
2
4
y ′ ( t ) = t −11/3
3
4
=
3 11
3 t
b.
f ′( x) = 0
(π
2
is a constant!)
T = sin −1 ( x ) ⋅sin −1 ( x )
dT
dx
=
=
=
d.
g are never linear.
d
sin ( x ) ) ⋅ sin ( x ) + sin ( x ) + ( sin ( x ) )
(
dx
dx
d
−1
1
1 − x2
−1
⋅ sin −1 ( x ) + sin −1 ( x ) ⋅
−1
1
1 − x2
2sin −1 ( x )
1 − x2
w = 3ln ( x ) − x
dw
1
1
= 3⋅ −
dx
x 2 x
x
3 2
1
= ⋅ −
⋅
x 2 2 x x
=
−1
e.
z=
4 x8
−2 x 2
= − 2 x6 ; x ≠ 0
dz
= − 12 x 5 ; x ≠ 0
dx
6− x
2x
Practice Test 2 – version B Key|1
Mr. Simonds’ MTH 251 Class
f.
1
3
z = ex − x
e
7
d
d
3 ⋅ ex − 3⋅ ( ex )
dz 1 x dx ( )
dx
= e −
x 2
dx 7
(e )
=
z (α ) = 25α
g.
z ′ (α ) = ln ( 25 ) ⋅ 25α
ex 0 ⋅ ex − 3⋅ ex
−
2
7
(ex )
ex ex 3 7
⋅ + ⋅
7 ex ex 7
e 2 x + 21
=
7 ex
=
h.
y = ⎡⎣sin ( 2θ ) ⎤⎦
2
= ⎡⎣ 2sin (θ ) cos (θ ) ⎤⎦
2
= 4sin (θ ) cos (θ ) sin (θ ) cos (θ )
dy
d
d
=
( 4sin (θ ) ) ⋅ cos (θ ) ⋅sin (θ ) ⋅ cos (θ ) + 4sin (θ ) ⋅ dθ ( cos (θ ) ) ⋅sin (θ ) ⋅ cos (θ )
dθ dθ
d
d
+ 4sin (θ ) ⋅ cos (θ ) ⋅ ( sin (θ ) ) ⋅ cos (θ ) + 4sin (θ ) ⋅ cos (θ ) ⋅ sin (θ ) ⋅ ( cos (θ ) )
dθ
dθ
= 4cos (θ ) ⋅ cos (θ ) ⋅ sin (θ ) ⋅ cos (θ ) + 4sin (θ ) ⋅ ( − sin (θ ) ) ⋅ sin (θ ) ⋅ cos (θ )
+ 4sin (θ ) ⋅ cos (θ ) ⋅ cos (θ ) ⋅ cos (θ ) + 4sin (θ ) ⋅ cos (θ ) ⋅ sin (θ ) ⋅ ( − sin (θ ) )
= 8sin (θ ) cos (θ ) − 8sin 3 (θ ) cos (θ )
3
= 8sin (θ ) cos (θ ) ( cos 2 (θ ) − sin 2 (θ ) )
i.
1
ex
d
d
1) ⋅ e x − 1⋅ ( e x )
(
dr
dx
= 0 + dx
2
x
dx
(e )
r =1+
=
0 ⋅ e x − 1⋅ e x
=−
3.
(e )
x 2
j.
d x
d
e ) ⋅ ( e x + 1) − e x ⋅ ( e x + 1)
(
dx
u ( x ) = dx
2
x
( e + 1)
=
=
e x ⋅ ( e x + 1) − e x ⋅ e x
(e
x
+ 1)
2
ex
(e
x
+ 1)
2
1
ex
f ( 4 ) = − 56 , so the tangent line passes through the point ( 4, − 56 ) .
f ′ ( x ) = 2 − 3 x 2 ⇒ f ′ ( 4 ) = − 46 ; so the slope of the tangent line is −46 .
Putting the pieces together we conclude that the equation of the tangent line to f at 4 is
y = − 46 x + 128 .
2|Practice Test 2 – version B Key
Mr. Simonds’ MTH 251 Class
4. Using Figure 2, p (1) = P′ (1) = 0 , so the tangent line to
p at 1 passes through the point (1,0 ) .
Using Figure 3, p ′ (1) = 2 , so the slope of the tangent line to p at 1 is 2 .
p at 1 is, thus, y = 2 x − 2 .
Using basic algebra, the equation of the tangent line to
5. The velocity function is v ( t ) = p′ ( t ) = cos ( t ) .
The acceleration function is a ( t ) = v′ ( t ) = − sin ( t ) .
⎛π ⎞
⎛ 3π
⎞
, 2 π ⎟ , so that is when
, π ⎟ and ⎜
⎝2 ⎠
⎝ 2
⎠
Over ( 0, 2 π ) these functions share a common sign on ⎜
the speed of the object is increasing.
6.
Figure 5K:
7.
f′
Figure 6K:
F
f is the first derivative of g .
8. We can determine the intervals over which g is increasing by determining the intervals where
g ′ is positive.
Since g ′′′ ( x ) = 0 at all values of x , the value of
g ′′ ( x ) is constant,;
specifically, g ′′ ( x ) = − 5 at all values of x . This means that g ′ has a constant slope of −5 .
Since we also know that g ′ passes through the point
( 3, − 25 ) , the formula for
g ′ must be
g ′ ( x ) = − 5 x − 10 .
g ′ ( x ) > 0 ⇒ − 5 x − 10 > 0 ⇒ − 5 x > 10 ⇒ x < − 2
Consequently, g is increasing over the interval ( −∞, − 2 ) .
Practice Test 2 – version B Key|3
Mr. Simonds’ MTH 251 Class
9.
x = − 5 x = −1 x = − 5
Figure 9K:
g′
Figure 10K:
4|Practice Test 2 – version B Key
G