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Math Problems
Algebra 1
Multiplication with rational exponents
d
d
x
(
1
x
12
)
=
\frac{d}{d x}\left(\frac{1}{x^{12}}\right)=
d
x
d
(
x
12
1
)
=
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Let
h
(
x
)
=
1
x
11
h(x)=\frac{1}{x^{11}}
h
(
x
)
=
x
11
1
.
\newline
h
′
(
x
)
=
h^{\prime}(x)=
h
′
(
x
)
=
Get tutor help
Let
g
(
x
)
=
1
x
10
g(x)=\frac{1}{x^{10}}
g
(
x
)
=
x
10
1
.
\newline
g
′
(
x
)
=
g^{\prime}(x)=
g
′
(
x
)
=
Get tutor help
What is the value of
d
d
x
(
x
3
)
\frac{d}{d x}\left(\sqrt{x^{3}}\right)
d
x
d
(
x
3
)
at
x
=
25
x=25
x
=
25
?
Get tutor help
Let
y
=
x
2
+
2
x
4
−
5
x
y=\frac{x^{2}+2 x}{4-5 x}
y
=
4
−
5
x
x
2
+
2
x
.
\newline
What is the value of
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at
x
=
2
?
x=2 ?
x
=
2
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
4
3
-\frac{4}{3}
−
3
4
\newline
(B)
1
9
\frac{1}{9}
9
1
\newline
(c)
−
6
5
-\frac{6}{5}
−
5
6
\newline
(D)
−
2
3
-\frac{2}{3}
−
3
2
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Let
f
(
x
)
=
x
sin
(
x
)
f(x)=\frac{\sqrt{x}}{\sin (x)}
f
(
x
)
=
s
i
n
(
x
)
x
.
\newline
Find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
1
2
x
cos
(
x
)
\frac{1}{2 \sqrt{x} \cos (x)}
2
x
c
o
s
(
x
)
1
\newline
(B)
cos
(
x
)
2
x
\frac{\cos (x)}{2 \sqrt{x}}
2
x
c
o
s
(
x
)
\newline
(C)
sin
(
x
)
−
2
x
cos
(
x
)
2
x
sin
2
(
x
)
\frac{\sin (x)-2 x \cos (x)}{2 \sqrt{x} \sin ^{2}(x)}
2
x
s
i
n
2
(
x
)
s
i
n
(
x
)
−
2
x
c
o
s
(
x
)
\newline
(D)
sin
(
x
)
+
2
x
cos
(
x
)
2
x
sin
2
(
x
)
\frac{\sin (x)+2 x \cos (x)}{2 \sqrt{x} \sin ^{2}(x)}
2
x
s
i
n
2
(
x
)
s
i
n
(
x
)
+
2
x
c
o
s
(
x
)
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What is the average value of
x
\sqrt{x}
x
on the interval
[
5
,
12
]
[5,12]
[
5
,
12
]
?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
21
⋅
(
1
2
3
−
5
3
)
\frac{2}{21} \cdot\left(\sqrt{12^{3}}-\sqrt{5^{3}}\right)
21
2
⋅
(
1
2
3
−
5
3
)
\newline
(B)
2
21
⋅
(
1
2
2
3
−
5
2
3
)
\frac{2}{21} \cdot\left(\sqrt[3]{12^{2}}-\sqrt[3]{5^{2}}\right)
21
2
⋅
(
3
1
2
2
−
3
5
2
)
\newline
(C)
1
2
⋅
(
12
−
5
)
\frac{1}{2} \cdot(\sqrt{12}-\sqrt{5})
2
1
⋅
(
12
−
5
)
\newline
(D)
1
2
⋅
(
12
+
5
)
\frac{1}{2} \cdot(\sqrt{12}+\sqrt{5})
2
1
⋅
(
12
+
5
)
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What is the average value of
x
3
\sqrt[3]{x}
3
x
on the interval
−
5
≤
x
≤
9
-5 \leq x \leq 9
−
5
≤
x
≤
9
?
\newline
Choose
1
1
1
answer:
\newline
(A)
3
56
⋅
(
9
3
4
−
5
3
4
)
\frac{3}{56} \cdot\left(\sqrt[4]{9^{3}}-\sqrt[4]{5^{3}}\right)
56
3
⋅
(
4
9
3
−
4
5
3
)
\newline
(B)
3
56
⋅
(
9
4
3
−
5
4
3
)
\frac{3}{56} \cdot\left(\sqrt[3]{9^{4}}-\sqrt[3]{5^{4}}\right)
56
3
⋅
(
3
9
4
−
3
5
4
)
\newline
(C)
1
2
⋅
(
9
3
−
−
5
3
)
\frac{1}{2} \cdot(\sqrt[3]{9}-\sqrt[3]{-5})
2
1
⋅
(
3
9
−
3
−
5
)
\newline
(D)
1
2
⋅
(
9
3
+
−
5
3
)
\frac{1}{2} \cdot(\sqrt[3]{9}+\sqrt[3]{-5})
2
1
⋅
(
3
9
+
3
−
5
)
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Let
g
(
x
)
=
x
e
3
x
g(x)=x e^{3 x}
g
(
x
)
=
x
e
3
x
.
\newline
What is the absolute minimum value of
g
g
g
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
3
e
-\frac{1}{3 e}
−
3
e
1
\newline
(B)
3
e
3
\frac{3}{e^{3}}
e
3
3
\newline
(C)
−
1
e
3
-\frac{1}{e^{3}}
−
e
3
1
\newline
(D)
g
g
g
has no minimum value
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Let
g
(
x
)
=
ln
(
x
)
x
g(x)=\frac{\ln (x)}{x}
g
(
x
)
=
x
l
n
(
x
)
.
\newline
What is the absolute maximum value of
g
g
g
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
e
-e
−
e
\newline
(B)
1
e
\frac{1}{e}
e
1
\newline
(C)
−
1
e
2
-\frac{1}{e^{2}}
−
e
2
1
\newline
(D)
g
g
g
has no maximum value
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Let
f
(
x
)
=
x
ln
(
x
)
f(x)=x \ln (x)
f
(
x
)
=
x
ln
(
x
)
.
\newline
What is the absolute minimum value of
f
f
f
?
\newline
Choose
1
1
1
answer:
\newline
(A)
0
0
0
\newline
(B)
1
e
\frac{1}{e}
e
1
\newline
(C)
−
1
e
-\frac{1}{e}
−
e
1
\newline
(D)
f
f
f
has no minimum value
Get tutor help
Find
d
2
d
x
2
[
ln
(
2
x
+
4
)
]
\frac{d^{2}}{d x^{2}}[\ln (2 x+4)]
d
x
2
d
2
[
ln
(
2
x
+
4
)]
.
Get tutor help
Let
g
(
x
)
=
−
5
x
g(x)=-\frac{5}{x}
g
(
x
)
=
−
x
5
.
\newline
Find
g
′
′
(
x
)
g^{\prime \prime}(x)
g
′′
(
x
)
.
\newline
g
′
′
(
x
)
=
g^{\prime \prime}(x)=
g
′′
(
x
)
=
Get tutor help
Let
g
(
x
)
=
−
8
sin
(
3
x
2
+
1
)
g(x)=-8 \sin \left(\frac{3 x}{2}+1\right)
g
(
x
)
=
−
8
sin
(
2
3
x
+
1
)
.
\newline
Find
g
′
′
(
x
)
g^{\prime \prime}(x)
g
′′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
8
(
3
x
2
+
1
)
2
⋅
sin
(
3
x
2
+
1
)
8\left(\frac{3 x}{2}+1\right)^{2} \cdot \sin \left(\frac{3 x}{2}+1\right)
8
(
2
3
x
+
1
)
2
⋅
sin
(
2
3
x
+
1
)
\newline
(B)
32
9
sin
(
3
x
2
+
1
)
\frac{32}{9} \sin \left(\frac{3 x}{2}+1\right)
9
32
sin
(
2
3
x
+
1
)
\newline
(C)
18
sin
(
3
x
2
+
1
)
18 \sin \left(\frac{3 x}{2}+1\right)
18
sin
(
2
3
x
+
1
)
\newline
(D)
−
12
cos
(
3
x
2
+
1
)
-12 \cos \left(\frac{3 x}{2}+1\right)
−
12
cos
(
2
3
x
+
1
)
Get tutor help
Let
g
(
x
)
=
ln
(
x
4
)
g(x)=\ln \left(x^{4}\right)
g
(
x
)
=
ln
(
x
4
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
4
x
\frac{4}{x}
x
4
\newline
(B)
1
ln
(
x
)
+
4
x
3
\frac{1}{\ln (x)}+4 x^{3}
l
n
(
x
)
1
+
4
x
3
\newline
(C)
1
x
4
\frac{1}{x^{4}}
x
4
1
\newline
(D)
4
ln
(
x
3
)
4 \ln \left(x^{3}\right)
4
ln
(
x
3
)
Get tutor help
Find
d
d
x
(
e
1
x
)
\frac{d}{d x}\left(e^{\frac{1}{x}}\right)
d
x
d
(
e
x
1
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
1
x
⋅
e
1
x
−
1
\frac{1}{x} \cdot e^{\frac{1}{x}-1}
x
1
⋅
e
x
1
−
1
\newline
(B)
−
e
1
x
-e^{\frac{1}{x}}
−
e
x
1
\newline
(C)
−
e
1
x
x
2
-\frac{e^{\frac{1}{x}}}{x^{2}}
−
x
2
e
x
1
\newline
(D)
e
1
x
x
\frac{e^{\frac{1}{x}}}{x}
x
e
x
1
Get tutor help
Let
g
(
x
)
=
ln
(
x
4
)
g(x)=\ln \left(x^{4}\right)
g
(
x
)
=
ln
(
x
4
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
4
x
\frac{4}{x}
x
4
\newline
(B)
1
x
4
\frac{1}{x^{4}}
x
4
1
\newline
(C)
4
ln
(
x
3
)
4 \ln \left(x^{3}\right)
4
ln
(
x
3
)
\newline
(D)
1
ln
(
x
)
+
4
x
3
\frac{1}{\ln (x)}+4 x^{3}
l
n
(
x
)
1
+
4
x
3
Get tutor help
Find
d
d
x
(
e
1
x
)
\frac{d}{d x}\left(e^{\frac{1}{x}}\right)
d
x
d
(
e
x
1
)
\newline
Choose
1
1
1
answer:
\newline
(A)
−
e
1
x
x
2
-\frac{e^{\frac{1}{x}}}{x^{2}}
−
x
2
e
x
1
\newline
(B)
1
x
⋅
e
1
x
−
1
\frac{1}{x} \cdot e^{\frac{1}{x}-1}
x
1
⋅
e
x
1
−
1
\newline
(C)
−
e
1
x
-e^{\frac{1}{x}}
−
e
x
1
\newline
(D)
e
1
x
x
\frac{e^{\frac{1}{x}}}{x}
x
e
x
1
Get tutor help
g
(
x
)
=
{
ln
(
x
)
for
0
<
x
≤
2
x
2
ln
(
2
)
for
x
>
2
g(x)=\left\{\begin{array}{ll} \ln (x) & \text { for } 0<x \leq 2 \\ x^{2} \ln (2) & \text { for } x>2 \end{array}\right.
g
(
x
)
=
{
ln
(
x
)
x
2
ln
(
2
)
for
0
<
x
≤
2
for
x
>
2
\newline
Find
lim
x
→
2
g
(
x
)
\lim _{x \rightarrow 2} g(x)
lim
x
→
2
g
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
ln
(
2
)
\ln (2)
ln
(
2
)
\newline
(B)
4
4
4
\newline
(C)
4
⋅
ln
(
2
)
4 \cdot \ln (2)
4
⋅
ln
(
2
)
\newline
(D) The limit doesn't exist.
Get tutor help
Simplify.
\newline
1
1
4
⋅
1
1
4
1^{\frac{1}{4}} \cdot 1^{\frac{1}{4}}
1
4
1
⋅
1
4
1
\newline
______
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