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Let’s check out your problem:
f
(
x
)
=
−
5
x
−
9
g
(
x
)
=
2
x
2
g
(
f
(
0
)
)
=
?
\begin{array}{l}f(x)=-5 x-9 \\ g(x)=2 x^{2} \\ g(f(0))=?\end{array}
f
(
x
)
=
−
5
x
−
9
g
(
x
)
=
2
x
2
g
(
f
(
0
))
=
?
View step-by-step help
Home
Math Problems
Calculus
Find higher derivatives of rational and radical functions
Full solution
Q.
f
(
x
)
=
−
5
x
−
9
g
(
x
)
=
2
x
2
g
(
f
(
0
)
)
=
?
\begin{array}{l}f(x)=-5 x-9 \\ g(x)=2 x^{2} \\ g(f(0))=?\end{array}
f
(
x
)
=
−
5
x
−
9
g
(
x
)
=
2
x
2
g
(
f
(
0
))
=
?
Find
f
(
0
)
f(0)
f
(
0
)
:
First, we need to find
f
(
0
)
f(0)
f
(
0
)
by plugging
0
0
0
into the function
f
(
x
)
f(x)
f
(
x
)
.
\newline
f
(
0
)
=
−
5
(
0
)
−
9
f(0) = -5(0) - 9
f
(
0
)
=
−
5
(
0
)
−
9
\newline
f
(
0
)
=
−
9
f(0) = -9
f
(
0
)
=
−
9
Calculate
g
(
f
(
0
)
)
g(f(0))
g
(
f
(
0
))
:
Now we plug
f
(
0
)
f(0)
f
(
0
)
into
g
(
x
)
g(x)
g
(
x
)
to find
g
(
f
(
0
)
)
g(f(0))
g
(
f
(
0
))
.
g
(
f
(
0
)
)
=
g
(
−
9
)
g(f(0)) = g(-9)
g
(
f
(
0
))
=
g
(
−
9
)
g
(
−
9
)
=
2
(
−
9
)
2
g(-9) = 2(-9)^2
g
(
−
9
)
=
2
(
−
9
)
2
Calculate
g
(
−
9
)
g(-9)
g
(
−
9
)
:
Calculate the square of
−
9
-9
−
9
and then multiply by
2
2
2
.\(\newline\)
(
−
9
)
2
=
81
(-9)^2 = 81
(
−
9
)
2
=
81
\(\newline\)
2
×
81
=
162
2 \times 81 = 162
2
×
81
=
162
\(\newline\)
g
(
−
9
)
=
162
g(-9) = 162
g
(
−
9
)
=
162
More problems from Find higher derivatives of rational and radical functions
Question
f
′
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12
e
x
and
f
(
4
)
=
−
16
+
12
e
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f
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0
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=
\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
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n
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n
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Posted 2 months ago
Question
The second derivative of the function
f
f
f
is defined by
f
′
′
(
x
)
=
x
2
−
5
cos
(
2
x
)
f^{\prime \prime}(x)=x^{2}-5 \cos (2 x)
f
′′
(
x
)
=
x
2
−
5
cos
(
2
x
)
for
−
2.5
<
x
<
3.5
-2.5<x<3.5
−
2.5
<
x
<
3.5
. Find the
x
x
x
-values, if any, in the given domain where the function
f
f
f
has an inflection point. You may use a calculator and round all values to
3
3
3
decimal places.
\newline
Answer:
x
=
x=
x
=
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Posted 2 months ago
Question
f
′
(
x
)
=
12
e
x
and
f
(
4
)
=
−
16
+
12
e
4
.
f
(
0
)
=
\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
f
′
(
x
)
=
12
e
x
and
f
(
4
)
=
−
16
+
12
e
4
.
f
(
0
)
=
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Posted 2 months ago
Question
x
2
−
6
x
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=
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x
2
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Posted 2 months ago
Question
=
x
2
+
4
=
(
g
+
h
)
(
x
)
\begin{array}{l} =x^{2}+4 \\ =(g+h)(x) \end{array}
=
x
2
+
4
=
(
g
+
h
)
(
x
)
\newline
9
9
9
.
\newline
h
(
x
)
=
x
2
+
x
g
(
x
)
=
3
x
+
5
Find
(
h
⋅
g
)
(
x
)
\begin{array}{l} h(x)=x^{2}+x \\ g(x)=3 x+5 \\ \text { Find }(h \cdot g)(x) \end{array}
h
(
x
)
=
x
2
+
x
g
(
x
)
=
3
x
+
5
Find
(
h
⋅
g
)
(
x
)
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Posted 2 months ago
Question
Factor completely:
\newline
8
c
3
+
27
m
6
8 c^{3}+27 m^{6}
8
c
3
+
27
m
6
\newline
Answer:
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Posted 2 months ago