Question
The graph of
on the interval
is shown above. Which of the following could be the graph of
?
A 
B 
C 
D 
▶️Answer/Explanation
Ans:A
The graph of
indicates that f
is increasing from x
to x=2, then decreasing from x
to x=3, and then increasing from x=3 to x=5. Therefore, the graph of
should be positive from x=0 to x=2, negative from
to
, and positive from x
to
. This graph is the only one that has this behavior, so it could be the graph of
. Some other features of the graph of
support this conclusion. Since f is not differentiable at
, the graph of
should not be defined at
. Since f
has a local minimum at
and is differentiable there,
should equal
. This graph is consistent with those observations.
Question
The equation \(y=2e^{6x}-5 \) is a particular solution to which of the following differential equations?
A y′−6y−30=0
B 2y′−12y+5=0
C y′′−5y′−6y=0
D y′′−2y′+y+5=0
▶️Answer/Explanation
Ans:A
Correct. One way to verify that a function is a solution to a differential equation is to check that the function and its derivatives satisfy the differential equation. The differential equation in this option involves y and y′. The correct derivative must be computed and the algebra correctly done to verify that the differential equation is satisfied.
\(y{}’=12e^{6x}\)
\(y{}’-6y-30=12e^{6x}-6(2e^{6x}-5)-30=0\)
Question
A particle moves along the x-axis so that its position at time t is given by \(x(t)=t^{2}-6t+5\).For what value of t is the velocity of the particle zero?
A 1
B 2
C 3
D 4
E 5
▶️Answer/Explanation
Ans:C
Question
An object moves along a straight line so that at any time t , 0≤t≤9, its position is given by \(x(t)=7+6t-t^{2}\). For what value of t is the object at rest?
A t=3
B t=6
C \( t=\frac{13}{2}\)
D t=7
▶️Answer/Explanation
Ans:A
The object is at rest when its velocity equals 0. Because the object’s velocity is given by x′(t)=6−2t, it is at rest at t=3.