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Let’s check out your problem:
Use the following function rule to find
f
(
2
)
f(2)
f
(
2
)
.
\newline
f
(
x
)
=
2
x
2
f(x) = 2x^2
f
(
x
)
=
2
x
2
\newline
f
(
2
)
=
f(2) =
f
(
2
)
=
_____
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Math Problems
Grade 8
Evaluate a nonlinear function
Full solution
Q.
Use the following function rule to find
f
(
2
)
f(2)
f
(
2
)
.
\newline
f
(
x
)
=
2
x
2
f(x) = 2x^2
f
(
x
)
=
2
x
2
\newline
f
(
2
)
=
f(2) =
f
(
2
)
=
_____
Substitute
x
x
x
with
2
2
2
:
Step
1
1
1
: Substitute
x
x
x
with
2
2
2
in the function rule.
\newline
Calculation:
f
(
2
)
=
2
(
2
)
2
f(2) = 2(2)^2
f
(
2
)
=
2
(
2
)
2
\newline
Math error check:
Calculate
(
2
)
2
(2)^2
(
2
)
2
:
Step
2
2
2
: Calculate
(
2
)
2
(2)^2
(
2
)
2
.
\newline
Calculation:
(
2
)
2
=
4
(2)^2 = 4
(
2
)
2
=
4
\newline
Math error check:
Multiply
2
2
2
by
4
4
4
:
Step
3
3
3
: Multiply
2
2
2
by
4
4
4
.
\newline
Calculation:
2
×
4
=
8
2 \times 4 = 8
2
×
4
=
8
\newline
Math error check:
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
)
=
Get tutor help
Posted 9 months ago
Question
If
y
2
=
2
x
2
+
y
y^{2}=2 x^{2}+y
y
2
=
2
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
)
=
Get tutor help
Posted 9 months ago
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
)
=
Get tutor help
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
)
=
Get tutor help
Posted 9 months ago
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
)
=
Get tutor help
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
)
=
Get tutor help
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
)
=
Get tutor help
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
)
=
Get tutor help
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
)
=
Get tutor help
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
=
Get tutor help
Posted 9 months ago
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