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Math Problems
Calculus
Find derivatives of logarithmic functions
f
(
t
)
=
{
−
64
t
,
t
=
8
14
−
t
,
t
=
10
t
2
−
3
t
+
2
,
t
≠
8
,
10
f
(
2
)
=
□
\begin{array}{l}f(t)=\left\{\begin{array}{ll}-\frac{64}{t} & , \quad t=8 \\ 14-t & , \quad t=10 \\ t^{2}-3 t+2 & , \quad t \neq 8,10\end{array}\right. \\ f(2)=\square\end{array}
f
(
t
)
=
⎩
⎨
⎧
−
t
64
14
−
t
t
2
−
3
t
+
2
,
t
=
8
,
t
=
10
,
t
=
8
,
10
f
(
2
)
=
□
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Classify the following numbers as either rational or irrational:
\newline
7
8
,
16
,
π
\frac{7}{8}, \sqrt{16}, \pi
8
7
,
16
,
π
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Simplify:
\newline
$
20
−
(
$
1.47
+
$
8
)
\$20-(\$1.47+\$8)
$20
−
(
$1.47
+
$8
)
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Find
y
′
y'
y
′
if
y
=
ln
(
4
+
x
2
x
)
.
y=\ln \left(\frac{\sqrt{4+x^{2}}}{x}\right).
y
=
ln
(
x
4
+
x
2
)
.
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Find the particular solution,
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
, to the differential equation
d
y
d
x
=
cos
x
y
\frac{d y}{d x}=\frac{\cos x}{y}
d
x
d
y
=
y
c
o
s
x
given
f
(
3
π
2
)
=
−
1
f\left(\frac{3 \pi}{2}\right)=-1
f
(
2
3
π
)
=
−
1
and then find
f
(
π
2
)
f\left(\frac{\pi}{2}\right)
f
(
2
π
)
.
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What is the particular solution to the differential equation
d
y
d
x
=
y
x
\frac{d y}{d x}=\frac{y}{x}
d
x
d
y
=
x
y
with the initial condition
y
(
e
2
)
=
−
1
y\left(e^{2}\right)=-1
y
(
e
2
)
=
−
1
?
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Question
\newline
What is the particular solution to the differential equation
d
y
d
x
=
1
+
y
x
\frac{d y}{d x}=\frac{1+y}{x}
d
x
d
y
=
x
1
+
y
with the initial condition
y
(
e
)
=
1
?
y(e)=1 ?
y
(
e
)
=
1
?
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Solve for the exact value of
x
x
x
.
\newline
log
6
(
9
x
)
−
log
6
(
4
)
=
2
\log_{6}(9x)-\log_{6}(4)=2
lo
g
6
(
9
x
)
−
lo
g
6
(
4
)
=
2
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What is
lim
x
→
∞
x
2
−
4
2
+
x
−
4
x
2
?
\lim _{x \rightarrow \infty} \frac{x^{2}-4}{2+x-4 x^{2}} ?
lim
x
→
∞
2
+
x
−
4
x
2
x
2
−
4
?
\newline
(A)
−
2
-2
−
2
\newline
(B)
−
1
4
-\frac{1}{4}
−
4
1
\newline
(C)
1
2
\frac{1}{2}
2
1
\newline
(D)
1
1
1
\newline
(E) The limit does not exist.
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Find the derivative of
f
(
x
)
f(x)
f
(
x
)
.
f
(
x
)
=
x
+
3
f(x) = \sqrt{x+3}
f
(
x
)
=
x
+
3
\newline
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
______
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Find the following values of the function
\newline
{
f
(
x
)
=
{
x
+
2
,
x
≤
5
6
−
x
,
x
>
5
\begin{cases} f(x) = \begin{cases} x+2, & x \leq 5 \ 6-x, & x > 5 \end{cases} \end{cases}
{
f
(
x
)
=
{
x
+
2
,
x
≤
5
6
−
x
,
x
>
5
,
\newline
f
(
2
)
=
□
f(2)=\square
f
(
2
)
=
□
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Solve for
x
x
x
:
\newline
x
4
−
3
=
2
\frac{x}{4}-3=2
4
x
−
3
=
2
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Solve for
t
t
t
.
\newline
4
3
=
t
7
\frac{4}{3}=\frac{t}{7}
3
4
=
7
t
\newline
t
=
□
t = \square
t
=
□
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Find
lim
h
→
0
2
tan
(
π
3
+
h
)
−
2
tan
(
π
3
)
h
\lim _{h \rightarrow 0} \frac{2 \tan \left(\frac{\pi}{3}+h\right)-2 \tan \left(\frac{\pi}{3}\right)}{h}
lim
h
→
0
h
2
t
a
n
(
3
π
+
h
)
−
2
t
a
n
(
3
π
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
\newline
(B)
2
2
2
\newline
(C)
8
8
8
\newline
(D) The limit doesn't exist
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Find
lim
h
→
0
5
log
(
2
+
h
)
−
5
log
(
2
)
h
\lim _{h \rightarrow 0} \frac{5 \log (2+h)-5 \log (2)}{h}
lim
h
→
0
h
5
l
o
g
(
2
+
h
)
−
5
l
o
g
(
2
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
5
2
\frac{5}{2}
2
5
\newline
(B)
5
2
ln
(
10
)
\frac{5}{2 \ln (10)}
2
l
n
(
10
)
5
\newline
(C)
5
log
(
2
)
5 \log (2)
5
lo
g
(
2
)
\newline
(D) The limit doesn't exist
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Find
lim
h
→
0
3
ln
(
e
+
h
)
−
3
ln
(
e
)
h
\lim _{h \rightarrow 0} \frac{3 \ln (e+h)-3 \ln (e)}{h}
lim
h
→
0
h
3
l
n
(
e
+
h
)
−
3
l
n
(
e
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
1
e
\frac{1}{e}
e
1
\newline
(B)
3
e
\frac{3}{e}
e
3
\newline
(c)
e
e
e
\newline
(D) The limit doesn't exist
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How does
f
(
t
)
=
2
t
f(t) = 2^t
f
(
t
)
=
2
t
change over the interval from
t
=
5
t = 5
t
=
5
to
t
=
6
t = 6
t
=
6
?
\newline
Choices:
\newline
(A)
f
(
t
)
f(t)
f
(
t
)
increases by a factor of
2
2
2
\newline
(B)
f
(
t
)
f(t)
f
(
t
)
decreases by a factor of
2
2
2
\newline
(C)
f
(
t
)
f(t)
f
(
t
)
increases by
200
%
200\%
200%
\newline
(D)
f
(
t
)
f(t)
f
(
t
)
increases by
2
2
2
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How does
g
(
t
)
=
2
t
g(t) = 2^t
g
(
t
)
=
2
t
change over the interval from
t
=
7
t = 7
t
=
7
to
t
=
8
t = 8
t
=
8
?
\newline
Choices:
\newline
(A)
g
(
t
)
g(t)
g
(
t
)
increases by
200
%
200\%
200%
\newline
(B)
g
(
t
)
g(t)
g
(
t
)
decreases by a factor of
2
2
2
\newline
(C)
g
(
t
)
g(t)
g
(
t
)
increases by a factor of
2
2
2
\newline
(D)
g
(
t
)
g(t)
g
(
t
)
increases by
t
=
7
t = 7
t
=
7
0
0
0
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How does
h
(
t
)
=
8
t
h(t) = 8^t
h
(
t
)
=
8
t
change over the interval from
t
=
6
t = 6
t
=
6
to
t
=
7
t = 7
t
=
7
?
\newline
Choices:
\newline
(A)
h
(
t
)
h(t)
h
(
t
)
increases by
800
%
800\%
800%
\newline
(B)
h
(
t
)
h(t)
h
(
t
)
decreases by
8
8
8
\newline
(C)
h
(
t
)
h(t)
h
(
t
)
increases by
8
%
8\%
8%
\newline
(D)
h
(
t
)
h(t)
h
(
t
)
increases by a factor of
8
8
8
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How does
f
(
t
)
=
9
t
f(t) = 9^t
f
(
t
)
=
9
t
change over the interval from
t
=
7
t = 7
t
=
7
to
t
=
8
t = 8
t
=
8
?
\newline
Choices:
\newline
(A)
f
(
t
)
f(t)
f
(
t
)
increases by a factor of
9
9
9
\newline
(B)
f
(
t
)
f(t)
f
(
t
)
decreases by a factor of
9
9
9
\newline
(C)
f
(
t
)
f(t)
f
(
t
)
decreases by
9
9
9
\newline
(D)
f
(
t
)
f(t)
f
(
t
)
increases by
9
9
9
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If
x
=
log
(
cos
(
y
2
)
−
sin
(
y
2
)
cos
(
y
2
)
+
sin
(
y
2
)
)
tan
(
y
2
)
=
1
−
1
1
+
1
x=\log\left(\frac{\cos\left(\frac{y}{2}\right)-\sin\left(\frac{y}{2}\right)}{\cos\left(\frac{y}{2}\right)+\sin\left(\frac{y}{2}\right)}\right)\tan\left(\frac{y}{2}\right)=\sqrt{\frac{1-1}{1+1}}
x
=
lo
g
(
c
o
s
(
2
y
)
+
s
i
n
(
2
y
)
c
o
s
(
2
y
)
−
s
i
n
(
2
y
)
)
tan
(
2
y
)
=
1
+
1
1
−
1
. Then
(
v
i
)
(vi)
(
v
i
)
,
o
o
o
, has the value
\newline
(A)
1
2
\frac{1}{2}
2
1
\newline
(B)
−
1
2
-\frac{1}{2}
−
2
1
\newline
(C)
1
4
\frac{1}{4}
4
1
\newline
(D)
−
1
4
-\frac{1}{4}
−
4
1
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Define limit of a function. If
x
2
+
y
2
+
3
x
y
=
7
x^{2}+y^{2}+3xy=7
x
2
+
y
2
+
3
x
y
=
7
, then find
(
d
y
)
/
(
d
x
)
(dy)/(dx)
(
d
y
)
/
(
d
x
)
If
y
=
cos
x
y=\cos x
y
=
cos
x
, then S.T.
y
1
2
+
y
2
=
1
y_{1}^{2}+y^{2}=1
y
1
2
+
y
2
=
1
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Solve for a positive value of
x
x
x
.
\newline
log
x
(
128
)
=
7
\log _{x}(128)=7
lo
g
x
(
128
)
=
7
\newline
Answer:
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Solve for a positive value of
x
x
x
.
\newline
log
x
(
16
)
=
2
\log _{x}(16)=2
lo
g
x
(
16
)
=
2
\newline
Answer:
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Evaluate the limit.
\newline
lim
x
→
0
sin
(
8
x
)
−
x
2
cos
(
6
x
)
x
\lim _{x \rightarrow 0} \frac{\operatorname{sin}(8 x)-x^{2} \cos \left(\frac{6}{x}\right)}{x}
lim
x
→
0
x
sin
(
8
x
)
−
x
2
c
o
s
(
x
6
)
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Find the derivative of
f
(
x
)
=
cos
(
2
ln
(
−
4
x
−
3
)
)
f(x)=\cos (2 \ln (-4 x-3))
f
(
x
)
=
cos
(
2
ln
(
−
4
x
−
3
))
.
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what could be possible 'one's' digits of the square roots of following number
1908
\newline 1908
1908
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Solve for
x
x
x
.
\newline
5
x
5
=
−
57
5
\frac{\sqrt{5}x}{5}=-\frac{57}{5}
5
5
x
=
−
5
57
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ln
(
x
)
+
ln
(
y
)
−
ln
(
x
y
)
\ln(x) + \ln(y) - \ln(xy)
ln
(
x
)
+
ln
(
y
)
−
ln
(
x
y
)
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Solve for
k
k
k
.
\newline
4
3
=
11
k
k
=
□
\begin{array}{l} \frac{4}{3}=\frac{11}{k} \\ k=\square \end{array}
3
4
=
k
11
k
=
□
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Find the derivative of
y
=
(
9
x
2
−
2
)
sec
(
x
)
y=\left(9 x^{2}-2\right)^{\sec (x)}
y
=
(
9
x
2
−
2
)
s
e
c
(
x
)
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Find the value of
f
(
−
1
)
+
f
(
2
)
−
f
(
4
)
f(-1)+f(2)-f(4)
f
(
−
1
)
+
f
(
2
)
−
f
(
4
)
, where
\newline
f
(
x
)
=
{
2
x
−
4
for
x
≥
4
x
for
0
≤
x
<
4
−
2
for
x
<
0
f(x)=\left\{\begin{array}{lll} \sqrt{2 x-4} & \text { for } x \geq 4 \\ x & \text { for } 0 \leq x<4 \\ -2 & \text { for } x<0 \end{array}\right.
f
(
x
)
=
⎩
⎨
⎧
2
x
−
4
x
−
2
for
x
≥
4
for
0
≤
x
<
4
for
x
<
0
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22
22
22
)
log
3
8
≈
1.9
log
3
10
≈
2.1
log
3
11
≈
2.2
Find
log
3
100
\begin{array}{l}\log _{3} 8 \approx 1.9 \\ \log _{3} 10 \approx 2.1 \\ \log _{3} 11 \approx 2.2 \\ \text { Find } \log _{3} 100\end{array}
lo
g
3
8
≈
1.9
lo
g
3
10
≈
2.1
lo
g
3
11
≈
2.2
Find
lo
g
3
100
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The horsepower,
H
H
H
produced by a truck engine is proportional to the cube of the truck's speed,
S
S
S
.
\newline
Calculate the percentage increase in horsepower that is needed to double the speed.
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lg
x
+
54
log
x
10
=
15
\lg x+54 \log _{x} 10=15
l
g
x
+
54
lo
g
x
10
=
15
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g
(
x
)
=
−
log
5
(
−
x
−
2
)
+
3
g(x)=-\log _{5}(-x-2)+3
g
(
x
)
=
−
lo
g
5
(
−
x
−
2
)
+
3
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Find the derivative of
f
(
x
)
f(x)
f
(
x
)
.
\newline
f
(
x
)
=
x
−
1
f(x)=\sqrt{x-1}
f
(
x
)
=
x
−
1
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g
(
x
)
=
−
x
2
4
+
7
g(x)=-\frac{x^{2}}{4}+7
g
(
x
)
=
−
4
x
2
+
7
\newline
What is the average rate of change of
g
g
g
over the interval
[
−
2
,
4
]
[-2,4]
[
−
2
,
4
]
?
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Find the derivative of the function
f
(
x
)
=
ln
(
x
2
−
3
x
+
2
)
f(x)=\ln(x^{2}-3x+2)
f
(
x
)
=
ln
(
x
2
−
3
x
+
2
)
.
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Solve for the exact value of
x
x
x
.
\newline
log
6
(
6
x
)
+
2
log
6
(
4
)
=
1
\log _{6}(6 x)+2 \log _{6}(4)=1
lo
g
6
(
6
x
)
+
2
lo
g
6
(
4
)
=
1
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
2
(
2
x
)
+
3
log
2
(
3
)
=
0
\log _{2}(2 x)+3 \log _{2}(3)=0
lo
g
2
(
2
x
)
+
3
lo
g
2
(
3
)
=
0
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
2
(
8
x
)
−
3
log
2
(
6
)
=
0
\log _{2}(8 x)-3 \log _{2}(6)=0
lo
g
2
(
8
x
)
−
3
lo
g
2
(
6
)
=
0
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
3
(
9
x
)
−
2
log
3
(
3
)
=
2
\log _{3}(9 x)-2 \log _{3}(3)=2
lo
g
3
(
9
x
)
−
2
lo
g
3
(
3
)
=
2
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
4
(
9
x
)
−
2
log
4
(
7
)
=
1
\log _{4}(9 x)-2 \log _{4}(7)=1
lo
g
4
(
9
x
)
−
2
lo
g
4
(
7
)
=
1
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
6
(
8
x
)
−
3
log
6
(
3
)
=
0
\log _{6}(8 x)-3 \log _{6}(3)=0
lo
g
6
(
8
x
)
−
3
lo
g
6
(
3
)
=
0
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
7
(
6
x
)
+
4
log
7
(
3
)
=
3
\log _{7}(6 x)+4 \log _{7}(3)=3
lo
g
7
(
6
x
)
+
4
lo
g
7
(
3
)
=
3
\newline
Answer:
Get tutor help
Solve for the exact value of
x
x
x
.
\newline
log
2
(
8
x
)
−
2
log
2
(
6
)
=
3
\log _{2}(8 x)-2 \log _{2}(6)=3
lo
g
2
(
8
x
)
−
2
lo
g
2
(
6
)
=
3
\newline
Answer:
Get tutor help
Solve for the exact value of
x
x
x
.
\newline
log
7
(
8
x
)
+
2
log
7
(
8
)
=
2
\log _{7}(8 x)+2 \log _{7}(8)=2
lo
g
7
(
8
x
)
+
2
lo
g
7
(
8
)
=
2
\newline
Answer:
Get tutor help
Solve for the exact value of
x
x
x
.
\newline
log
6
(
5
x
)
−
3
log
6
(
5
)
=
1
\log _{6}(5 x)-3 \log _{6}(5)=1
lo
g
6
(
5
x
)
−
3
lo
g
6
(
5
)
=
1
\newline
Answer:
Get tutor help
Solve for the exact value of
x
x
x
.
\newline
log
4
(
6
x
)
+
2
log
4
(
9
)
=
1
\log _{4}(6 x)+2 \log _{4}(9)=1
lo
g
4
(
6
x
)
+
2
lo
g
4
(
9
)
=
1
\newline
Answer:
Get tutor help
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