My orders
How it works
Examples
Reviews
Blog
Homework Answers
Submit
Sign in
How it works
Examples
Reviews
Homework answers
Blog
Contact us
Submit
Question #131234
Use an appropriate local linear approximation to estimate the value of (122)^2/3
Expert's answer
f
(
x
)
=
x
2
3
f(x)=x^{\frac {2}{3}}
f
(
x
)
=
x
3
2
f
(
x
0
+
Δ
x
)
≈
f
(
x
0
)
+
d
[
f
(
x
0
)
]
f(x_0+\varDelta x)\approx f(x_0)+d[f(x_0)]
f
(
x
0
+
Δ
x
)
≈
f
(
x
0
)
+
d
[
f
(
x
0
)]
x
0
=
125
,
Δ
x
=
−
3
x_0=125, \varDelta x=-3
x
0
=
125
,
Δ
x
=
−
3
d
[
f
(
x
0
)
]
=
f
/
(
x
0
)
×
Δ
x
d[f(x_0)]=f^/(x_0)\times \varDelta x
d
[
f
(
x
0
)]
=
f
/
(
x
0
)
×
Δ
x
f
/
(
x
)
=
2
3
x
1
3
f^/(x)=\frac {2}{3x^{\frac{1}{3}}}
f
/
(
x
)
=
3
x
3
1
2
d
[
f
(
x
0
)
]
=
f
/
(
x
0
)
×
Δ
x
=
2
15
×
(
−
3
)
=
−
0.4
d[f(x_0)]=f^/(x_0)\times \varDelta x=\frac {2}{15} \times(-3)=-0.4
d
[
f
(
x
0
)]
=
f
/
(
x
0
)
×
Δ
x
=
15
2
×
(
−
3
)
=
−
0.4
f
(
x
0
+
Δ
x
)
≈
f
(
x
0
)
+
d
[
f
(
x
0
)
]
≈
25
−
0.4
=
24.6
f(x_0+\varDelta x)\approx f(x_0)+d[f(x_0)] \\ \approx 25-0.4=24.6
f
(
x
0
+
Δ
x
)
≈
f
(
x
0
)
+
d
[
f
(
x
0
)]
≈
25
−
0.4
=
24.6
Our fields of expertise
Programming
Math
Engineering
Economics
Physics
LATEST TUTORIALS
APPROVED BY CLIENTS
Thanks for the assistance,,, your service is great
#340795
on Jan 2024
Read all reviews >>