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Math Problems
Calculus
Find derivatives of using multiple formulae
Given the function
y
=
(
−
6
−
10
x
−
1
−
x
)
(
−
7
−
7
x
3
)
y=\left(-6-10 x^{-1}-x\right)\left(-7-7 x^{3}\right)
y
=
(
−
6
−
10
x
−
1
−
x
)
(
−
7
−
7
x
3
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
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2
π
∫
−
3
0
−
x
3
12
1
+
x
4
16
2 \pi \int_{-\sqrt{3}}^{0}-\frac{x^{3}}{12} \sqrt{1+\frac{x^{4}}{16}}
2
π
∫
−
3
0
−
12
x
3
1
+
16
x
4
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If
x
=
t
2
−
1
x=t^{2}-1
x
=
t
2
−
1
and
y
=
ln
t
y=\ln t
y
=
ln
t
, what is
d
2
y
d
x
2
\frac{d^{2} y}{d x^{2}}
d
x
2
d
2
y
in terms of
t
t
t
?
\newline
(a)
−
1
2
t
4
-\frac{1}{2 t^{4}}
−
2
t
4
1
\newline
b)
1
2
t
4
\frac{1}{2 t^{4}}
2
t
4
1
\newline
c)
−
1
x
3
-\frac{1}{x^{3}}
−
x
3
1
\newline
d)
−
1
−
2
t
2
-\frac{1}{-2 t^{2}}
−
−
2
t
2
1
\newline
e)
1
2
t
2
\frac{1}{2 t^{2}}
2
t
2
1
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V
=
π
∫
−
1
2
[
x
+
2
]
2
−
[
x
2
]
2
d
x
V=\pi \int_{-1}^{2}[x+2]^{2}-\left[x^{2}\right]^{2} d x
V
=
π
∫
−
1
2
[
x
+
2
]
2
−
[
x
2
]
2
d
x
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Simplify.
\newline
(
20
m
9
n
−
5
p
7
15
m
−
4
n
−
2
p
−
7
)
−
1
\left(\frac{20 m^{9} n^{-5} p^{7}}{15 m^{-4} n^{-2} p^{-7}}\right)^{-1}
(
15
m
−
4
n
−
2
p
−
7
20
m
9
n
−
5
p
7
)
−
1
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Find the limit as
𝑥
𝑥
x
approaches 10 of
lim
x
→
10
x
2
−
100
x
−
10
\lim_{x \to 10}\frac{x^{2}-100}{x-10}
lim
x
→
10
x
−
10
x
2
−
100
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Using the definition of the derivative, find
f
′
(
x
)
f'(x)
f
′
(
x
)
.
\newline
f
(
x
)
=
−
x
2
+
6
x
−
7
f(x)=-x^{2}+6x-7
f
(
x
)
=
−
x
2
+
6
x
−
7
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Which expressions are equivalent to
b
9
5
\sqrt[5]{b^{9}}
5
b
9
?
\newline
Choose all answers that apply:
\newline
(
A
)
(A)
(
A
)
(
b
5
)
1
9
(b^{5})^{\frac{1}{9}}
(
b
5
)
9
1
\newline
(
B
)
(B)
(
B
)
b
5
9
b^{\frac{5}{9}}
b
9
5
\newline
(
C
)
(C)
(
C
)
b
9
5
b^{\frac{9}{5}}
b
5
9
\newline
(
D
)
(D)
(
D
)
None of the
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Which expressions are equivalent to
(
d
1
8
)
5
(d^{\frac{1}{8}})^5
(
d
8
1
)
5
?
\newline
Choose all answers that apply:
\newline
(
A
)
(A)
(
A
)
(
d
5
)
1
8
(d^5)^{\frac{1}{8}}
(
d
5
)
8
1
\newline
(
B
)
(B)
(
B
)
(
d
5
)
8
(d^5)^8
(
d
5
)
8
\newline
(
C
)
(C)
(
C
)
d
8
5
\sqrt[8]{d}^5
8
d
5
\newline
(
D
)
(D)
(
D
)
None of the above
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For any integer
m
m
m
and
n
n
n
, which of the following expression is equivalent to
2
5
m
n
25^{mn}
2
5
mn
?
\newline
A.
(
5
n
)
2
m
(5^{n})^{2m}
(
5
n
)
2
m
\newline
B.
5
m
⋅
5
n
5^{m}\cdot5^{n}
5
m
⋅
5
n
\newline
C.
2
5
m
⋅
2
5
n
25^{m}\cdot25^{n}
2
5
m
⋅
2
5
n
\newline
D.
2
5
m
+
2
5
n
25^{m}+25^{n}
2
5
m
+
2
5
n
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A cardboard cereal box is
9
in.
9\,\text{in.}
9
in.
by
2.5
in.
2.5\,\text{in.}
2.5
in.
by
10
in.
10\,\text{in.}
10
in.
What is the minimum amount of cardboard needed to create this cereal box?
\newline
A
)
$
275
in.
2
A) \$275\,\text{in.}^2
A
)
$275
in.
2
\newline
B
)
$
230
in.
2
B) \$230\,\text{in.}^2
B
)
$230
in.
2
\newline
C
)
$
225
in.
2
C) \$225\,\text{in.}^2
C
)
$225
in.
2
\newline
D
)
$
180
in.
2
D) \$180\,\text{in.}^2
D
)
$180
in.
2
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Numeric expression with the resulting value
\newline
(
4
4
×
3
2
×
4
−
2
)
/
(
4
×
3
−
1
×
3
0
)
(4^{4}\times3^{2}\times4^{-2})/(4\times3^{-1}\times3^{0})
(
4
4
×
3
2
×
4
−
2
)
/
(
4
×
3
−
1
×
3
0
)
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Simplify.
(
x
4
7
−
8
)
−
7
\left(\frac{x^{4}}{7^{-8}}\right)^{-7}
(
7
−
8
x
4
)
−
7
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If
u
=
e
x
y
+
e
y
z
+
e
z
x
u = e^{\frac{x}{y}} + e^{\frac{y}{z}} + e^{\frac{z}{x}}
u
=
e
y
x
+
e
z
y
+
e
x
z
then show that
x
(
∂
u
∂
x
)
+
y
(
∂
u
∂
y
)
+
z
(
∂
u
∂
z
)
=
0
x\left(\frac{\partial u}{\partial x}\right) + y\left(\frac{\partial u}{\partial y}\right) + z\left(\frac{\partial u}{\partial z}\right) = 0
x
(
∂
x
∂
u
)
+
y
(
∂
y
∂
u
)
+
z
(
∂
z
∂
u
)
=
0
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Find
k
′
(
x
)
k'(x)
k
′
(
x
)
if
k
(
x
)
=
e
x
(
−
x
3
5
−
x
−
5
4
)
k(x)=e^{x}(-x^{\frac{3}{5}}-x^{-\frac{5}{4}})
k
(
x
)
=
e
x
(
−
x
5
3
−
x
−
4
5
)
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(Adding and Subtracting with Scientific Notation LC)
\newline
Find the difference of
5.482
×
1
0
12
5.482 \times 10^{12}
5.482
×
1
0
12
and
3.2
×
1
0
12
3.2 \times 10^{12}
3.2
×
1
0
12
.
\newline
A.
5.162
×
1
0
12
5.162 \times 10^{12}
5.162
×
1
0
12
\newline
B.
2.282
×
1
0
12
2.282 \times 10^{12}
2.282
×
1
0
12
\newline
C.
5.162
×
1
0
0
5.162 \times 10^{0}
5.162
×
1
0
0
\newline
D.
2.282
×
1
0
0
2.282 \times 10^{0}
2.282
×
1
0
0
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A curve is defined by the parametric equations
x
(
t
)
=
7
e
−
8
t
x(t) = 7e^{-8t}
x
(
t
)
=
7
e
−
8
t
and
y
(
t
)
=
−
7
sin
(
−
10
t
)
y(t) = -7\sin(-10t)
y
(
t
)
=
−
7
sin
(
−
10
t
)
. Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
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Find
k
′
(
x
)
k'(x)
k
′
(
x
)
if
k
(
x
)
=
e
x
(
4
5
x
3
+
3
4
x
−
2
)
k(x)=e^{x}\left(\frac{4}{5}x^{3}+\frac{3}{4}x^{-2}\right)
k
(
x
)
=
e
x
(
5
4
x
3
+
4
3
x
−
2
)
.
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Find
k
′
(
x
)
k'(x)
k
′
(
x
)
if
\newline
k
(
x
)
=
e
x
(
−
3
5
x
−
4
5
+
1
5
x
−
3
2
)
k(x)=e^{x}\left(-\frac{3}{5}x^{-\frac{4}{5}}+\frac{1}{5}x^{-\frac{3}{2}}\right)
k
(
x
)
=
e
x
(
−
5
3
x
−
5
4
+
5
1
x
−
2
3
)
.
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Simplify:
5
c
(
3
c
2
)
3
5c(3c^{2})^{3}
5
c
(
3
c
2
)
3
\newline
A.
45
c
6
45c^{6}
45
c
6
\newline
B.
135
c
6
135c^{6}
135
c
6
\newline
C.
45
c
7
45c^{7}
45
c
7
\newline
D.
135
c
7
135c^{7}
135
c
7
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If
u
=
e
x
y
+
e
y
z
+
e
z
x
u=e^{\frac{x}{y}}+e^{\frac{y}{z}}+e^{\frac{z}{x}}
u
=
e
y
x
+
e
z
y
+
e
x
z
then show that
x
∂
u
∂
x
+
y
∂
u
∂
y
+
z
∂
u
∂
z
=
0
x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=0
x
∂
x
∂
u
+
y
∂
y
∂
u
+
z
∂
z
∂
u
=
0
\newline
e
x
y
(
x
y
+
x
y
)
+
e
y
z
(
y
x
+
2
y
)
+
e
z
x
(
x
1
z
+
z
)
e^{\frac{x}{y}}\left(\frac{x}{y}+xy\right)+e^{\frac{y}{z}}\left(\frac{y}{x}+2y\right)+e^{\frac{z}{x}}(x_{1}z+z)
e
y
x
(
y
x
+
x
y
)
+
e
z
y
(
x
y
+
2
y
)
+
e
x
z
(
x
1
z
+
z
)
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z
=
log
(
x
−
2
)
3
y
−
64
−
x
2
−
y
2
z = \frac{\log(\sqrt{x} - 2)}{3y} - \sqrt{64 - x^{2} - y^{2}}
z
=
3
y
l
o
g
(
x
−
2
)
−
64
−
x
2
−
y
2
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Prove.
(
a
−
b
)
2
=
a
2
−
2
⋅
a
⋅
b
+
b
2
(a-b)^{2}=a^{2}-2 \cdot a \cdot b+b^{2}
(
a
−
b
)
2
=
a
2
−
2
⋅
a
⋅
b
+
b
2
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Evaluate the summation below.
\newline
∑
i
=
0
3
(
−
5
i
2
−
6
i
)
\sum_{i=0}^{3}\left(-5 i^{2}-6 i\right)
i
=
0
∑
3
(
−
5
i
2
−
6
i
)
\newline
Answer:
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Determine
h
′
(
1
)
h'(1)
h
′
(
1
)
if
h
(
x
)
=
f
(
g
(
2
−
x
2
)
)
h(x)=f(\sqrt{g(2-x^2)})
h
(
x
)
=
f
(
g
(
2
−
x
2
)
)
: if
g
(
1
)
=
4
g(1)=4
g
(
1
)
=
4
,
f
′
(
2
)
=
2
f'(2)=2
f
′
(
2
)
=
2
,
g
(
1
)
=
1
4
g(1)=\frac{1}{4}
g
(
1
)
=
4
1
,
g
′
(
1
)
=
1
2
g'(1)=\frac{1}{2}
g
′
(
1
)
=
2
1
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What is the value of
A
A
A
when we rewrite
(
5
6
)
x
\left(\frac{5}{6}\right)^{x}
(
6
5
)
x
as
A
−
8
x
A^{-8x}
A
−
8
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
(
5
6
)
1
8
A=\left(\frac{5}{6}\right)^{\frac{1}{8}}
A
=
(
6
5
)
8
1
\newline
(B)
A
=
(
5
6
)
−
1
8
A=\left(\frac{5}{6}\right)^{-\frac{1}{8}}
A
=
(
6
5
)
−
8
1
\newline
(C)
A
=
−
8
⋅
5
6
A=-8\cdot\frac{5}{6}
A
=
−
8
⋅
6
5
\newline
(D)
A
=
1
8
A=\frac{1}{8}
A
=
8
1
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Differentiate
y
=
e
4
x
tan
x
ln
x
y=\frac{\mathrm{e}^{4 x} \tan x}{\ln x}
y
=
l
n
x
e
4
x
t
a
n
x
with respect to
x
x
x
.
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y
′
′
′
+
6
y
′
′
+
11
y
′
+
6
y
=
0
y^{\prime \prime \prime}+6 y^{\prime \prime}+11 y^{\prime}+6 y=0
y
′′′
+
6
y
′′
+
11
y
′
+
6
y
=
0
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1
1
1
. Differentiate the following functions:
\newline
(a)
f
(
x
)
=
e
x
2
+
2
e
x
+
x
e
2
+
x
e
2
+
x
e
x
f(x)=e x^{2}+2 e^{x}+x e^{2}+x^{e^{2}}+x e^{x}
f
(
x
)
=
e
x
2
+
2
e
x
+
x
e
2
+
x
e
2
+
x
e
x
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What is the value of
A
A
A
when we rewrite
(
5
6
)
x
\left(\frac{5}{6}\right)^{x}
(
6
5
)
x
as
A
−
8
x
A^{-8 x}
A
−
8
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
(
5
6
)
1
8
A=\left(\frac{5}{6}\right)^{\frac{1}{8}}
A
=
(
6
5
)
8
1
\newline
(B)
A
=
(
5
6
)
−
1
8
A=\left(\frac{5}{6}\right)^{-\frac{1}{8}}
A
=
(
6
5
)
−
8
1
\newline
(C)
A
=
−
8
⋅
5
6
A=-8 \cdot \frac{5}{6}
A
=
−
8
⋅
6
5
\newline
(D)
A
=
1
8
A=\frac{1}{8}
A
=
8
1
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∫
2
x
3
+
7
x
2
+
7
x
+
9
(
x
2
+
x
+
1
)
(
x
2
+
x
+
2
)
d
x
\int \frac{2 x^{3}+7 x^{2}+7 x+9}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)} d x
∫
(
x
2
+
x
+
1
)
(
x
2
+
x
+
2
)
2
x
3
+
7
x
2
+
7
x
+
9
d
x
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(i) A square park has a length
50
m
50 \mathrm{~m}
50
m
. Its perimeter is
\newline
(a)
2500
m
2500 \mathrm{~m}
2500
m
\newline
(b)
100
m
100 \mathrm{~m}
100
m
\newline
(c)
200
m
2
200 \mathrm{~m}^{2}
200
m
2
\newline
(d)
200
m
200 \mathrm{~m}
200
m
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Simplify. Your answer should be in proper scientific notation:
\newline
(
2
×
1
0
3
)
(
4
×
1
0
5
)
(
8
×
1
0
3
)
\left(2 \times 10^{3}\right)\left(4 \times 10^{5}\right)\left(8 \times 10^{3}\right)
(
2
×
1
0
3
)
(
4
×
1
0
5
)
(
8
×
1
0
3
)
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Fully simplify.
\newline
7
x
3
y
(
−
16
x
5
y
2
)
7 x^{3} y\left(-16 x^{5} y^{2}\right)
7
x
3
y
(
−
16
x
5
y
2
)
\newline
Answer:
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Combine like terms.
\newline
−
5
+
3
+
2
y
2
−
5
x
2
−
y
2
+
3
x
2
+
3
x
2
-5+3+2 y^{2}-5 x^{2}-y^{2}+3 x^{2}+3 x^{2}
−
5
+
3
+
2
y
2
−
5
x
2
−
y
2
+
3
x
2
+
3
x
2
\newline
Answer:
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Combine like terms.
\newline
−
7
x
3
+
6
x
2
+
x
2
−
7
x
2
−
7
y
3
+
7
y
3
+
4
x
3
-7 x^{3}+6 x^{2}+x^{2}-7 x^{2}-7 y^{3}+7 y^{3}+4 x^{3}
−
7
x
3
+
6
x
2
+
x
2
−
7
x
2
−
7
y
3
+
7
y
3
+
4
x
3
\newline
Answer:
Get tutor help
Combine like terms.
\newline
−
7
x
2
−
5
−
y
−
x
2
+
3
y
+
2
+
5
-7 x^{2}-5-y-x^{2}+3 y+2+5
−
7
x
2
−
5
−
y
−
x
2
+
3
y
+
2
+
5
\newline
Answer:
Get tutor help
Combine like terms.
\newline
−
5
y
2
+
x
2
−
3
x
2
+
6
x
2
−
5
y
2
+
4
x
3
+
3
x
3
-5 y^{2}+x^{2}-3 x^{2}+6 x^{2}-5 y^{2}+4 x^{3}+3 x^{3}
−
5
y
2
+
x
2
−
3
x
2
+
6
x
2
−
5
y
2
+
4
x
3
+
3
x
3
\newline
Answer:
Get tutor help
Simplify the expression completely if possible.
\newline
5
x
4
+
15
x
3
x
2
−
6
x
−
27
\frac{5 x^{4}+15 x^{3}}{x^{2}-6 x-27}
x
2
−
6
x
−
27
5
x
4
+
15
x
3
\newline
Answer:
Get tutor help
Simplify the expression completely if possible.
\newline
x
2
−
11
x
+
18
x
4
−
x
3
\frac{x^{2}-11 x+18}{x^{4}-x^{3}}
x
4
−
x
3
x
2
−
11
x
+
18
\newline
Answer:
Get tutor help
For the function
f
(
x
)
=
(
x
+
4
)
7
f(x)=\frac{(x+4)}{7}
f
(
x
)
=
7
(
x
+
4
)
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
7
(
x
+
4
)
f^{-1}(x)=7(x+4)
f
−
1
(
x
)
=
7
(
x
+
4
)
\newline
f
−
1
(
x
)
=
7
(
x
−
4
)
f^{-1}(x)=7(x-4)
f
−
1
(
x
)
=
7
(
x
−
4
)
\newline
f
−
1
(
x
)
=
x
7
−
4
f^{-1}(x)=\frac{x}{7}-4
f
−
1
(
x
)
=
7
x
−
4
\newline
f
−
1
(
x
)
=
7
x
−
4
f^{-1}(x)=7 x-4
f
−
1
(
x
)
=
7
x
−
4
Get tutor help
For the function
f
(
x
)
=
5
x
+
10
5
f(x)=5 \sqrt[5]{x+10}
f
(
x
)
=
5
5
x
+
10
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
x
5
5
−
10
f^{-1}(x)=\frac{x^{5}}{5}-10
f
−
1
(
x
)
=
5
x
5
−
10
\newline
f
−
1
(
x
)
=
(
x
5
)
5
−
10
f^{-1}(x)=\left(\frac{x}{5}\right)^{5}-10
f
−
1
(
x
)
=
(
5
x
)
5
−
10
\newline
f
−
1
(
x
)
=
(
x
5
−
10
)
5
f^{-1}(x)=\left(\frac{x}{5}-10\right)^{5}
f
−
1
(
x
)
=
(
5
x
−
10
)
5
\newline
f
−
1
(
x
)
=
(
x
−
10
)
5
5
f^{-1}(x)=\frac{(x-10)^{5}}{5}
f
−
1
(
x
)
=
5
(
x
−
10
)
5
Get tutor help
For the function
f
(
x
)
=
8
x
3
+
4
f(x)=8 \sqrt[3]{x}+4
f
(
x
)
=
8
3
x
+
4
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
(
x
8
)
3
−
4
f^{-1}(x)=\left(\frac{x}{8}\right)^{3}-4
f
−
1
(
x
)
=
(
8
x
)
3
−
4
\newline
f
−
1
(
x
)
=
(
x
−
4
8
)
3
f^{-1}(x)=\left(\frac{x-4}{8}\right)^{3}
f
−
1
(
x
)
=
(
8
x
−
4
)
3
\newline
f
−
1
(
x
)
=
x
3
8
−
4
f^{-1}(x)=\frac{x^{3}}{8}-4
f
−
1
(
x
)
=
8
x
3
−
4
\newline
f
−
1
(
x
)
=
(
x
8
−
4
)
3
f^{-1}(x)=\left(\frac{x}{8}-4\right)^{3}
f
−
1
(
x
)
=
(
8
x
−
4
)
3
Get tutor help
For the function
f
(
x
)
=
2
(
x
−
3
)
1
7
f(x)=2(x-3)^{\frac{1}{7}}
f
(
x
)
=
2
(
x
−
3
)
7
1
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
x
7
+
3
2
f^{-1}(x)=\frac{x^{7}+3}{2}
f
−
1
(
x
)
=
2
x
7
+
3
\newline
f
−
1
(
x
)
=
(
x
2
)
7
+
3
f^{-1}(x)=\left(\frac{x}{2}\right)^{7}+3
f
−
1
(
x
)
=
(
2
x
)
7
+
3
\newline
f
−
1
(
x
)
=
x
7
2
+
3
f^{-1}(x)=\frac{x^{7}}{2}+3
f
−
1
(
x
)
=
2
x
7
+
3
\newline
f
−
1
(
x
)
=
(
x
+
3
2
)
7
f^{-1}(x)=\left(\frac{x+3}{2}\right)^{7}
f
−
1
(
x
)
=
(
2
x
+
3
)
7
Get tutor help
For the function
f
(
x
)
=
x
3
−
8
5
f(x)=\frac{x^{3}-8}{5}
f
(
x
)
=
5
x
3
−
8
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
5
x
3
+
8
f^{-1}(x)=5 \sqrt[3]{x}+8
f
−
1
(
x
)
=
5
3
x
+
8
\newline
f
−
1
(
x
)
=
5
x
+
8
3
f^{-1}(x)=\sqrt[3]{5 x+8}
f
−
1
(
x
)
=
3
5
x
+
8
\newline
f
−
1
(
x
)
=
5
(
x
+
8
)
3
f^{-1}(x)=\sqrt[3]{5(x+8)}
f
−
1
(
x
)
=
3
5
(
x
+
8
)
\newline
f
−
1
(
x
)
=
5
x
+
8
3
f^{-1}(x)=5 \sqrt[3]{x+8}
f
−
1
(
x
)
=
5
3
x
+
8
Get tutor help
For the function
f
(
x
)
=
(
x
−
5
7
)
1
5
f(x)=\left(\frac{x-5}{7}\right)^{\frac{1}{5}}
f
(
x
)
=
(
7
x
−
5
)
5
1
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
7
x
5
+
5
f^{-1}(x)=7 x^{5}+5
f
−
1
(
x
)
=
7
x
5
+
5
\newline
f
−
1
(
x
)
=
(
7
x
+
5
)
5
f^{-1}(x)=(7 x+5)^{5}
f
−
1
(
x
)
=
(
7
x
+
5
)
5
\newline
f
−
1
(
x
)
=
7
(
x
+
5
)
5
f^{-1}(x)=7(x+5)^{5}
f
−
1
(
x
)
=
7
(
x
+
5
)
5
\newline
f
−
1
(
x
)
=
7
(
x
5
+
5
)
f^{-1}(x)=7\left(x^{5}+5\right)
f
−
1
(
x
)
=
7
(
x
5
+
5
)
Get tutor help
For the function
f
(
x
)
=
x
5
+
6
f(x)=x^{5}+6
f
(
x
)
=
x
5
+
6
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
x
5
+
6
f^{-1}(x)=\sqrt[5]{x}+6
f
−
1
(
x
)
=
5
x
+
6
\newline
f
−
1
(
x
)
=
x
5
−
6
f^{-1}(x)=\sqrt[5]{x}-6
f
−
1
(
x
)
=
5
x
−
6
\newline
f
−
1
(
x
)
=
x
+
6
5
f^{-1}(x)=\sqrt[5]{x+6}
f
−
1
(
x
)
=
5
x
+
6
\newline
f
−
1
(
x
)
=
x
−
6
5
f^{-1}(x)=\sqrt[5]{x-6}
f
−
1
(
x
)
=
5
x
−
6
Get tutor help
For the function
f
(
x
)
=
x
5
−
10
f(x)=x^{5}-10
f
(
x
)
=
x
5
−
10
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
(
x
+
10
)
5
f^{-1}(x)=(x+10)^{5}
f
−
1
(
x
)
=
(
x
+
10
)
5
\newline
f
−
1
(
x
)
=
x
−
10
5
f^{-1}(x)=\sqrt[5]{x-10}
f
−
1
(
x
)
=
5
x
−
10
\newline
f
−
1
(
x
)
=
x
+
10
5
f^{-1}(x)=\sqrt[5]{x+10}
f
−
1
(
x
)
=
5
x
+
10
\newline
f
−
1
(
x
)
=
x
5
+
10
f^{-1}(x)=x^{5}+10
f
−
1
(
x
)
=
x
5
+
10
Get tutor help
For the function
f
(
x
)
=
x
1
7
+
8
f(x)=x^{\frac{1}{7}}+8
f
(
x
)
=
x
7
1
+
8
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
(
x
−
8
)
1
7
f^{-1}(x)=(x-8)^{\frac{1}{7}}
f
−
1
(
x
)
=
(
x
−
8
)
7
1
\newline
f
−
1
(
x
)
=
x
1
7
−
8
f^{-1}(x)=x^{\frac{1}{7}}-8
f
−
1
(
x
)
=
x
7
1
−
8
\newline
f
−
1
(
x
)
=
(
x
−
8
)
7
f^{-1}(x)=(x-8)^{7}
f
−
1
(
x
)
=
(
x
−
8
)
7
\newline
f
−
1
(
x
)
=
x
7
−
8
f^{-1}(x)=x^{7}-8
f
−
1
(
x
)
=
x
7
−
8
Get tutor help
For the function
f
(
x
)
=
7
(
x
−
6
)
f(x)=7(x-6)
f
(
x
)
=
7
(
x
−
6
)
, find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
.
\newline
f
−
1
(
x
)
=
7
(
x
+
6
)
f^{-1}(x)=7(x+6)
f
−
1
(
x
)
=
7
(
x
+
6
)
\newline
f
−
1
(
x
)
=
x
7
−
6
f^{-1}(x)=\frac{x}{7}-6
f
−
1
(
x
)
=
7
x
−
6
\newline
f
−
1
(
x
)
=
(
x
−
6
)
7
f^{-1}(x)=\frac{(x-6)}{7}
f
−
1
(
x
)
=
7
(
x
−
6
)
\newline
f
−
1
(
x
)
=
x
7
+
6
f^{-1}(x)=\frac{x}{7}+6
f
−
1
(
x
)
=
7
x
+
6
Get tutor help
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