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Let’s check out your problem:
Use the following function rule to find
f
(
−
6
)
f(-6)
f
(
−
6
)
.
\newline
f
(
x
)
=
−
10
∣
x
∣
+
4
f(x) = -10|x| + 4
f
(
x
)
=
−
10∣
x
∣
+
4
\newline
f
(
−
6
)
=
f(-6) =
f
(
−
6
)
=
_____
View step-by-step help
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Math Problems
Grade 8
Evaluate a nonlinear function
Full solution
Q.
Use the following function rule to find
f
(
−
6
)
f(-6)
f
(
−
6
)
.
\newline
f
(
x
)
=
−
10
∣
x
∣
+
4
f(x) = -10|x| + 4
f
(
x
)
=
−
10∣
x
∣
+
4
\newline
f
(
−
6
)
=
f(-6) =
f
(
−
6
)
=
_____
Substitute
−
6
-6
−
6
for
x
x
x
:
Step
1
1
1
: Substitute
–
6
–6
–6
for
x
x
x
in the function rule.
f
(
–
6
)
=
–
10
∣
–
6
∣
+
4
f(–6) = –10|–6| + 4
f
(
–6
)
=
–10∣–6∣
+
4
Calculate absolute value:
Step
2
2
2
: Calculate the absolute value of
–
6
–6
–6
.
∣
–
6
∣
=
6
|–6| = 6
∣–6∣
=
6
Multiply by
−
10
-10
−
10
:
Step
3
3
3
: Multiply
–
10
–10
–10
by the absolute value.
–
10
×
6
=
–
60
–10 \times 6 = –60
–10
×
6
=
–60
Add
4
4
4
:
Step
4
4
4
: Add
4
4
4
to the result from Step
3
3
3
.
\newline
−
60
+
4
=
−
56
-60 + 4 = -56
−
60
+
4
=
−
56
More problems from Evaluate a nonlinear function
Question
If
−
x
2
=
−
5
y
2
+
y
3
-x^{2}=-5 y^{2}+y^{3}
−
x
2
=
−
5
y
2
+
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
2
,
1
)
(2,1)
(
2
,
1
)
.
\newline
Answer:
d
y
d
x
∣
(
2
,
1
)
=
\left.\frac{d y}{d x}\right|_{(2,1)}=
d
x
d
y
∣
∣
(
2
,
1
)
=
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Question
If
y
2
=
2
x
2
+
y
y^{2}=2 x^{2}+y
y
2
=
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x
2
+
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
2
)
=
\left.\frac{d y}{d x}\right|_{(1,2)}=
d
x
d
y
∣
∣
(
1
,
2
)
=
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Question
If
−
3
−
3
x
2
−
5
y
=
−
y
3
−
y
2
-3-3 x^{2}-5 y=-y^{3}-y^{2}
−
3
−
3
x
2
−
5
y
=
−
y
3
−
y
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
1
,
−
2
)
(-1,-2)
(
−
1
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
−
1
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(-1,-2)}=
d
x
d
y
∣
∣
(
−
1
,
−
2
)
=
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Posted 9 months ago
Question
If
5
x
2
−
5
−
3
y
2
=
0
5 x^{2}-5-3 y^{2}=0
5
x
2
−
5
−
3
y
2
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
4
,
5
)
(4,5)
(
4
,
5
)
.
\newline
Answer:
d
y
d
x
∣
(
4
,
5
)
=
\left.\frac{d y}{d x}\right|_{(4,5)}=
d
x
d
y
∣
∣
(
4
,
5
)
=
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Question
If
−
2
x
2
+
y
2
+
y
3
+
3
x
=
0
-2 x^{2}+y^{2}+y^{3}+3 x=0
−
2
x
2
+
y
2
+
y
3
+
3
x
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
2
,
1
)
(2,1)
(
2
,
1
)
.
\newline
Answer:
d
y
d
x
∣
(
2
,
1
)
=
\left.\frac{d y}{d x}\right|_{(2,1)}=
d
x
d
y
∣
∣
(
2
,
1
)
=
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Posted 9 months ago
Question
If
y
2
+
3
x
3
=
2
y
y^{2}+3 x^{3}=2 y
y
2
+
3
x
3
=
2
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
1
,
3
)
(-1,3)
(
−
1
,
3
)
.
\newline
Answer:
d
y
d
x
∣
(
−
1
,
3
)
=
\left.\frac{d y}{d x}\right|_{(-1,3)}=
d
x
d
y
∣
∣
(
−
1
,
3
)
=
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Posted 8 months ago
Question
If
4
x
−
4
−
4
x
2
=
y
2
−
y
3
4 x-4-4 x^{2}=y^{2}-y^{3}
4
x
−
4
−
4
x
2
=
y
2
−
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
2
)
=
\left.\frac{d y}{d x}\right|_{(1,2)}=
d
x
d
y
∣
∣
(
1
,
2
)
=
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Posted 8 months ago
Question
If
0
=
4
+
5
x
3
−
y
2
0=4+5 x^{3}-y^{2}
0
=
4
+
5
x
3
−
y
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
−
3
)
(1,-3)
(
1
,
−
3
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
−
3
)
=
\left.\frac{d y}{d x}\right|_{(1,-3)}=
d
x
d
y
∣
∣
(
1
,
−
3
)
=
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Posted 8 months ago
Question
If
−
y
3
−
y
+
2
+
y
2
=
x
2
-y^{3}-y+2+y^{2}=x^{2}
−
y
3
−
y
+
2
+
y
2
=
x
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
4
,
−
2
)
(4,-2)
(
4
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
4
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(4,-2)}=
d
x
d
y
∣
∣
(
4
,
−
2
)
=
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Posted 8 months ago
Question
Given
y
=
−
2
sin
(
2
x
)
y=-2 \sin (2 x)
y
=
−
2
sin
(
2
x
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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Posted 9 months ago
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