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Math Problems
Algebra 1
Evaluate an exponential function
For all integers
x
>
0
x>0
x
>
0
, let
f
(
x
)
f(x)
f
(
x
)
be defined as
f
(
x
)
=
f
(
x
−
1
)
(
−
1
)
x
f(x)=\frac{f(x-1)}{(-1)^x}
f
(
x
)
=
(
−
1
)
x
f
(
x
−
1
)
. If
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
, which of the following statement is correct for the values of
f
(
x
)
f(x)
f
(
x
)
?
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Given the definitions of
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
below, find the value of
f
(
g
(
−
2
)
)
f(g(-2))
f
(
g
(
−
2
))
.
\newline
f
(
x
)
=
2
x
−
3
g
(
x
)
=
3
x
2
+
7
x
+
12
\begin{array}{l} f(x)=2 x-3 \\ g(x)=3 x^{2}+7 x+12 \end{array}
f
(
x
)
=
2
x
−
3
g
(
x
)
=
3
x
2
+
7
x
+
12
\newline
Answer:
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Given the definitions of
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
below, find the value of
f
(
g
(
0
)
)
f(g(0))
f
(
g
(
0
))
.
\newline
f
(
x
)
=
−
5
x
−
6
g
(
x
)
=
x
2
−
3
x
−
10
\begin{array}{l} f(x)=-5 x-6 \\ g(x)=x^{2}-3 x-10 \end{array}
f
(
x
)
=
−
5
x
−
6
g
(
x
)
=
x
2
−
3
x
−
10
\newline
Answer:
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Given the definitions of
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
below, find the value of
(
g
∘
f
)
(
0
)
(g \circ f)(0)
(
g
∘
f
)
(
0
)
.
\newline
f
(
x
)
=
−
x
−
4
g
(
x
)
=
x
2
−
4
x
−
15
\begin{array}{l} f(x)=-x-4 \\ g(x)=x^{2}-4 x-15 \end{array}
f
(
x
)
=
−
x
−
4
g
(
x
)
=
x
2
−
4
x
−
15
\newline
Answer:
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Use the following function rule to find
f
(
1
)
f(1)
f
(
1
)
.
\newline
f
(
x
)
=
(
x
+
12
)
2
f
(
1
)
=
□
\begin{array}{l} f(x)=(x+12)^{2} \\ f(1)=\square \end{array}
f
(
x
)
=
(
x
+
12
)
2
f
(
1
)
=
□
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- Let
f
f
f
be a function such that
f
(
−
1
)
=
3
f(-1)=3
f
(
−
1
)
=
3
and
f
′
(
−
1
)
=
5
f^{\prime}(-1)=5
f
′
(
−
1
)
=
5
.
\newline
- Let
g
g
g
be the function
g
(
x
)
=
1
x
g(x)=\frac{1}{x}
g
(
x
)
=
x
1
.
\newline
Let
F
F
F
be a function defined as
F
(
x
)
=
f
(
x
)
⋅
g
(
x
)
F(x)=f(x) \cdot g(x)
F
(
x
)
=
f
(
x
)
⋅
g
(
x
)
.
\newline
F
′
(
−
1
)
=
F^{\prime}(-1)=
F
′
(
−
1
)
=
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g
(
x
)
=
−
20
−
3
x
g(x)=-20-3x
g
(
x
)
=
−
20
−
3
x
\newline
h
(
x
)
=
(
1
2
)
x
h(x)=(\frac{1}{2})^x
h
(
x
)
=
(
2
1
)
x
\newline
Evaluate.
\newline
(
g
∘
h
)
(
−
2
)
=
(g \circ h)(-2) =
(
g
∘
h
)
(
−
2
)
=
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Write an explicit formula that represents the sequence defined by the following recursive formula:
\newline
a
1
=
80
and
a
n
=
−
1
4
a
n
−
1
a_{1}=80 \text { and } a_{n}=-\frac{1}{4} a_{n-1}
a
1
=
80
and
a
n
=
−
4
1
a
n
−
1
\newline
Answer:
a
n
=
a_{n}=
a
n
=
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Write an explicit formula that represents the sequence defined by the following recursive formula:
\newline
a
1
=
75
and
a
n
=
1
5
a
n
−
1
a_{1}=75 \text { and } a_{n}=\frac{1}{5} a_{n-1}
a
1
=
75
and
a
n
=
5
1
a
n
−
1
\newline
Answer:
a
n
=
a_{n}=
a
n
=
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Rewrite the expression as a product of four linear factors:
\newline
(
x
2
+
x
)
2
−
22
(
x
2
+
x
)
+
40
\left(x^{2}+x\right)^{2}-22\left(x^{2}+x\right)+40
(
x
2
+
x
)
2
−
22
(
x
2
+
x
)
+
40
\newline
Answer:
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Write an explicit formula for
a
n
a_{n}
a
n
, the
n
th
n^{\text {th }}
n
th
term of the sequence
17
,
27
,
37
,
…
17,27,37, \ldots
17
,
27
,
37
,
…
.
\newline
Answer:
a
n
=
a_{n}=
a
n
=
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For the following equation, evaluate
d
y
d
x
\frac{d y}{d x}
d
x
d
y
when
x
=
2
x=2
x
=
2
.
\newline
y
=
3
x
3
+
4
y=3 x^{3}+4
y
=
3
x
3
+
4
\newline
Answer:
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If
−
x
−
x
y
=
−
y
3
-x-x y=-y^{3}
−
x
−
x
y
=
−
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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If
2
x
y
−
2
y
=
−
3
x
3
2 x y-2 y=-3 x^{3}
2
x
y
−
2
y
=
−
3
x
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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If
−
2
y
2
−
x
=
2
y
3
+
x
3
-2 y^{2}-x=2 y^{3}+x^{3}
−
2
y
2
−
x
=
2
y
3
+
x
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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If
y
=
5
y
3
+
2
x
3
y=5 y^{3}+2 x^{3}
y
=
5
y
3
+
2
x
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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If
5
y
+
y
3
−
y
2
−
x
3
=
0
5 y+y^{3}-y^{2}-x^{3}=0
5
y
+
y
3
−
y
2
−
x
3
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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If
0
=
x
3
−
y
+
3
−
5
y
3
0=x^{3}-y+3-5 y^{3}
0
=
x
3
−
y
+
3
−
5
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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If
y
2
−
x
2
−
3
=
0
y^{2}-x^{2}-3=0
y
2
−
x
2
−
3
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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Let
f
(
x
)
=
x
−
2
f(x)=x-2
f
(
x
)
=
x
−
2
and
g
(
x
)
=
x
2
+
x
−
1
g(x)=x^{2}+x-1
g
(
x
)
=
x
2
+
x
−
1
. Find each of the following and simplify.
\newline
a)
(
f
g
)
(
x
)
=
(fg)(x)=
(
f
g
)
(
x
)
=
\newline
b)
(
f
g
)
(
−
1
)
=
(fg)(-1)=
(
f
g
)
(
−
1
)
=
\newline
c)
(
f
g
)
(
2
)
=
(fg)(2)=
(
f
g
)
(
2
)
=
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Find
lim
x
→
−
1
x
2
−
9
x
2
+
1
\lim _{x \rightarrow-1} \frac{x^{2}-9}{x^{2}+1}
lim
x
→
−
1
x
2
+
1
x
2
−
9
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
4
-4
−
4
\newline
(B)
−
5
-5
−
5
\newline
(C)
−
9
-9
−
9
\newline
(D) The limit doesn't exist
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Find
lim
x
→
−
3
−
5
(
x
+
3
)
2
\lim _{x \rightarrow-3} \frac{-5}{(x+3)^{2}}
lim
x
→
−
3
(
x
+
3
)
2
−
5
.
\newline
Choose
1
1
1
answer:
\newline
(A)
0
0
0
\newline
(B)
−
5
36
-\frac{5}{36}
−
36
5
\newline
(c)
−
5
6
-\frac{5}{6}
−
6
5
\newline
(D) The limit doesn't exist
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∑
n
=
0
1
(
n
+
6
)
=
\sum_{n=0}^{1}(n+6)=
∑
n
=
0
1
(
n
+
6
)
=
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Find the following trigonometric values.
\newline
Express your answers exactly.
\newline
cos
(
15
0
∘
)
=
sin
(
15
0
∘
)
=
\begin{array}{l} \cos \left(150^{\circ}\right)= \\ \sin \left(150^{\circ}\right)= \end{array}
cos
(
15
0
∘
)
=
sin
(
15
0
∘
)
=
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Use the following function rule to find
f
(
2
)
f(2)
f
(
2
)
.
\newline
f
(
x
)
=
11
⋅
(
9
)
x
+
8
f(x) = 11 \cdot (9)^x + 8
f
(
x
)
=
11
⋅
(
9
)
x
+
8
\newline
f
(
2
)
=
f(2) =
f
(
2
)
=
_____
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