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# 𝐷𝑎𝑑𝑎 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑓(𝑥) = −0.1𝑥 4 − 0.15𝑥 3 − 0.5𝑥 2 − 0.25𝑥 + 1.2.

## 𝑈𝑠𝑒 𝑙𝑎𝑠 𝑎𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑒 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑠

𝑓𝑖𝑛𝑖𝑡𝑎𝑠 ℎ𝑎𝑐𝑖𝑎 𝑎𝑑𝑒𝑙𝑎𝑛𝑡𝑒, ℎ𝑎𝑐𝑖𝑎 𝑎𝑡𝑟á𝑠 𝑦 𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑎𝑠 𝑝𝑎𝑟𝑎 𝑒𝑠𝑡𝑖𝑚𝑎𝑟 𝑙𝑎 𝑝𝑟𝑖𝑚𝑒𝑟𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎.

 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟐 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 − ℎ) = 𝑓(0) = −0.1(0)4 − 0.15(0)3 − 0.5(0)2 − 0.25(0) + 1.2 = 𝟏. 𝟐

𝑓(𝑥0 ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =

𝑓(0.5) − 𝑓(0)
𝑓´(0.5) =
0.5
0.925 − 1.2
𝑓´(0.5) = = −𝟎. 𝟓𝟓
0.5

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 + ℎ) = 𝑓(1) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2 = 𝟎. 𝟐

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =
2ℎ
𝑓(1) − 𝑓(0)
𝑓´(0.5) =
2(0.5)
0.2 − 1.2
𝑓´(0.5) = = −𝟏. 𝟎
2(0.5)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 ) = 𝑓(0.5) = −0.1(0.5)4 − 0.15(0.5)3 − 0.5(0.5)2 − 0.25(0.5) + 1.2 = 𝟎. 𝟗𝟐𝟓

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 )
𝑓´(𝑥0 ) =

𝑓(1) − 𝑓(0.5)
𝑓´(0.5) =
0.5
0.2 − 0.925
𝑓´(0.5) = = −𝟏. 𝟒𝟓
0.5
 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟐 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟐𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 − ℎ) = 𝑓(0.25) = −0.1(0.25)4 − 0.15(0.25)3 − 0.5(0.25)2 − 0.25(0.25) + 1.2 = 𝟏. 𝟏𝟎𝟑𝟓𝟏𝟓𝟔𝟐𝟓

𝑓(𝑥0 ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =

𝑓(0.5) − 𝑓(0.25)
𝑓´(0.5) =
0.5
0.925 − 1.103515625
𝑓´(0.5) = = −𝟎. 𝟕𝟏𝟒𝟎𝟔𝟐𝟓
0.25

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 + ℎ) = 𝑓(0.75) = −0.1(0.75)4 − 0.15(0.75)3 − 0.5(0.75)2 − 0.25(0.75) + 1.2 = 𝟎. 𝟔𝟑𝟔𝟑𝟐𝟖𝟏𝟐𝟓

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =
2ℎ
𝑓(0.75) − 𝑓(0.25)
𝑓´(0.5) =
2(0.25)
1.103515625 − 0.636328125
𝑓´(0.5) = = −𝟎. 𝟗𝟑𝟒𝟑𝟕𝟓
2(0.25)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 ) = 𝑓(0.5) = −0.1(0.5)4 − 0.15(0.5)3 − 0.5(0.5)2 − 0.25(0.5) + 1.2 = 𝟎. 𝟗𝟐𝟓

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 )
𝑓´(𝑥0 ) =

𝑓(0.75) − 𝑓(0.5)
𝑓´(0.5) =
0.25
0.636328125 − 0.925
𝑓´(0.5) = = −𝟏. 𝟏𝟓𝟒𝟒𝟕𝟓
0.25
 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒔𝒆𝒈𝒖𝒏𝒅𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟑 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

𝑓´(𝑥0 ) =
2ℎ

## 𝑓(−0.5) − 4𝑓(0) + 3𝑓(0.5)

𝑓´(0.5) =
2(0.5)
1.2125 − 4(1.2) + 3(0.925)
𝑓´(0.5) = = −𝟎. 𝟖𝟏𝟐𝟓
2(0.5)

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 + ℎ) = 𝑓(1) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2 = 𝟎. 𝟐

𝑓(𝑥0 + 2ℎ) = 𝑓(1.5) = −0.1(1.5)4 − 0.15(1.5)3 − 0.5(1.5)2 − 0.25(1.5) + 1.2 = −𝟏. 𝟑𝟏𝟐𝟓

## 𝑓(𝑥0 − 2ℎ) − 8𝑓(𝑥0 − ℎ) + 8𝑓(𝑥0 + ℎ) + 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
12ℎ
𝑓(−0.5) − 8𝑓(0) + 8𝑓(1) − 𝑓(1.5)
𝑓´(0.5) =
12(0.5)
1.2125 − 8(1.2) + 8(0.2) − (−1.3125)
𝑓´(0.5) = = −𝟎. 𝟗𝟏𝟐𝟓
12(0.5)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 + ℎ) = 𝑓(1) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2 = 𝟎. 𝟐

𝑓(𝑥0 + 2ℎ) = 𝑓(1.5) = −0.1(1.5)4 − 0.15(1.5)3 − 0.5(1.5)2 − 0.25(1.5) + 1.2 = −𝟏. 𝟑𝟏𝟐𝟓

## −3𝑓(𝑥0 ) + 4𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
2ℎ
−3𝑓(0.5) + 4𝑓(1) − 𝑓(1.5)
𝑓´(0.5) =
2(0.5)
−3(0.925) + 4(0.2) − (−1.3125)
𝑓´(0.5) = = −𝟎. 𝟔𝟔𝟐𝟓
2(0.5)
 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒔𝒆𝒈𝒖𝒏𝒅𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟑 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟐𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

𝑓´(𝑥0 ) =
2ℎ

## 𝑓(0) − 4𝑓(0.25) + 3𝑓(0.5)

𝑓´(0.5) =
2(0.25)
1.2 − 4(1.103515625) + 3(0.925)
𝑓´(0.5) = = −𝟎. 𝟖𝟕𝟖𝟏𝟐𝟓
2(0.25)

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 − 2ℎ) − 8𝑓(𝑥0 − ℎ) + 8𝑓(𝑥0 + ℎ) + 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
12ℎ
𝑓(0) − 8𝑓(0.25) + 8𝑓(0.75) − 𝑓(1)
𝑓´(0.5) =
12(0.25)
1.2 − 8(1.103515625) + 8(0.636328125) − 0.2
𝑓´(0.5) = = −𝟎. 𝟗𝟏𝟐𝟓
12(0.25)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## −3𝑓(𝑥0 ) + 4𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
2ℎ
−3𝑓(0.5) + 4𝑓(0.75) − 𝑓(1)
𝑓´(0.5) =
2(0.25)
−3(0.925) + 4(0.636328125) − 0.2
𝑓´(0.5) = = −𝟎. 𝟖𝟓𝟗𝟑𝟕𝟓
2(0.25)
The text passage contains a series of mathematical questions and their corresponding answers, all related to calculating the first derivative of a given function using different finite difference methods.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Está en la página 1de 4

# 𝐷𝑎𝑑𝑎 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑓(𝑥) = −0.1𝑥 4 − 0.15𝑥 3 − 0.5𝑥 2 − 0.25𝑥 + 1.2.

## 𝑈𝑠𝑒 𝑙𝑎𝑠 𝑎𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑒 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑠

𝑓𝑖𝑛𝑖𝑡𝑎𝑠 ℎ𝑎𝑐𝑖𝑎 𝑎𝑑𝑒𝑙𝑎𝑛𝑡𝑒, ℎ𝑎𝑐𝑖𝑎 𝑎𝑡𝑟á𝑠 𝑦 𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑎𝑠 𝑝𝑎𝑟𝑎 𝑒𝑠𝑡𝑖𝑚𝑎𝑟 𝑙𝑎 𝑝𝑟𝑖𝑚𝑒𝑟𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎.

 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟐 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 − ℎ) = 𝑓(0) = −0.1(0)4 − 0.15(0)3 − 0.5(0)2 − 0.25(0) + 1.2 = 𝟏. 𝟐

𝑓(𝑥0 ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =

𝑓(0.5) − 𝑓(0)
𝑓´(0.5) =
0.5
0.925 − 1.2
𝑓´(0.5) = = −𝟎. 𝟓𝟓
0.5

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 + ℎ) = 𝑓(1) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2 = 𝟎. 𝟐

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =
2ℎ
𝑓(1) − 𝑓(0)
𝑓´(0.5) =
2(0.5)
0.2 − 1.2
𝑓´(0.5) = = −𝟏. 𝟎
2(0.5)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 ) = 𝑓(0.5) = −0.1(0.5)4 − 0.15(0.5)3 − 0.5(0.5)2 − 0.25(0.5) + 1.2 = 𝟎. 𝟗𝟐𝟓

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 )
𝑓´(𝑥0 ) =

𝑓(1) − 𝑓(0.5)
𝑓´(0.5) =
0.5
0.2 − 0.925
𝑓´(0.5) = = −𝟏. 𝟒𝟓
0.5
 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟐 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟐𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 − ℎ) = 𝑓(0.25) = −0.1(0.25)4 − 0.15(0.25)3 − 0.5(0.25)2 − 0.25(0.25) + 1.2 = 𝟏. 𝟏𝟎𝟑𝟓𝟏𝟓𝟔𝟐𝟓

𝑓(𝑥0 ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =

𝑓(0.5) − 𝑓(0.25)
𝑓´(0.5) =
0.5
0.925 − 1.103515625
𝑓´(0.5) = = −𝟎. 𝟕𝟏𝟒𝟎𝟔𝟐𝟓
0.25

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 + ℎ) = 𝑓(0.75) = −0.1(0.75)4 − 0.15(0.75)3 − 0.5(0.75)2 − 0.25(0.75) + 1.2 = 𝟎. 𝟔𝟑𝟔𝟑𝟐𝟖𝟏𝟐𝟓

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 − ℎ)
𝑓´(𝑥0 ) =
2ℎ
𝑓(0.75) − 𝑓(0.25)
𝑓´(0.5) =
2(0.25)
1.103515625 − 0.636328125
𝑓´(0.5) = = −𝟎. 𝟗𝟑𝟒𝟑𝟕𝟓
2(0.25)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 ) = 𝑓(0.5) = −0.1(0.5)4 − 0.15(0.5)3 − 0.5(0.5)2 − 0.25(0.5) + 1.2 = 𝟎. 𝟗𝟐𝟓

𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 )
𝑓´(𝑥0 ) =

𝑓(0.75) − 𝑓(0.5)
𝑓´(0.5) =
0.25
0.636328125 − 0.925
𝑓´(0.5) = = −𝟏. 𝟏𝟓𝟒𝟒𝟕𝟓
0.25
 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒔𝒆𝒈𝒖𝒏𝒅𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟑 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

𝑓´(𝑥0 ) =
2ℎ

## 𝑓(−0.5) − 4𝑓(0) + 3𝑓(0.5)

𝑓´(0.5) =
2(0.5)
1.2125 − 4(1.2) + 3(0.925)
𝑓´(0.5) = = −𝟎. 𝟖𝟏𝟐𝟓
2(0.5)

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 + ℎ) = 𝑓(1) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2 = 𝟎. 𝟐

𝑓(𝑥0 + 2ℎ) = 𝑓(1.5) = −0.1(1.5)4 − 0.15(1.5)3 − 0.5(1.5)2 − 0.25(1.5) + 1.2 = −𝟏. 𝟑𝟏𝟐𝟓

## 𝑓(𝑥0 − 2ℎ) − 8𝑓(𝑥0 − ℎ) + 8𝑓(𝑥0 + ℎ) + 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
12ℎ
𝑓(−0.5) − 8𝑓(0) + 8𝑓(1) − 𝑓(1.5)
𝑓´(0.5) =
12(0.5)
1.2125 − 8(1.2) + 8(0.2) − (−1.3125)
𝑓´(0.5) = = −𝟎. 𝟗𝟏𝟐𝟓
12(0.5)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## 𝑓(𝑥0 + ℎ) = 𝑓(1) = −0.1(1)4 − 0.15(1)3 − 0.5(1)2 − 0.25(1) + 1.2 = 𝟎. 𝟐

𝑓(𝑥0 + 2ℎ) = 𝑓(1.5) = −0.1(1.5)4 − 0.15(1.5)3 − 0.5(1.5)2 − 0.25(1.5) + 1.2 = −𝟏. 𝟑𝟏𝟐𝟓

## −3𝑓(𝑥0 ) + 4𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
2ℎ
−3𝑓(0.5) + 4𝑓(1) − 𝑓(1.5)
𝑓´(0.5) =
2(0.5)
−3(0.925) + 4(0.2) − (−1.3125)
𝑓´(0.5) = = −𝟎. 𝟔𝟔𝟐𝟓
2(0.5)
 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒓 𝒑𝒐𝒓 𝒆𝒍 𝒎é𝒕𝒐𝒅𝒐 𝒅𝒆 𝒔𝒆𝒈𝒖𝒏𝒅𝒂𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒇𝒊𝒏𝒊𝒕𝒂𝒔 (𝟑 𝒑𝒖𝒏𝒕𝒐𝒔), 𝒍𝒂 𝒑𝒓𝒊𝒎𝒆𝒓𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒅𝒆 𝒇(𝒙) 𝒆𝒏
𝒙𝟎 = 𝟎. 𝟓 𝒚 𝒉 = 𝟎. 𝟐𝟓

𝑅𝑒𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

𝑓´(𝑥0 ) =
2ℎ

## 𝑓(0) − 4𝑓(0.25) + 3𝑓(0.5)

𝑓´(0.5) =
2(0.25)
1.2 − 4(1.103515625) + 3(0.925)
𝑓´(0.5) = = −𝟎. 𝟖𝟕𝟖𝟏𝟐𝟓
2(0.25)

𝐶𝑒𝑛𝑡𝑟𝑎𝑙:

## 𝑓(𝑥0 − 2ℎ) − 8𝑓(𝑥0 − ℎ) + 8𝑓(𝑥0 + ℎ) + 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
12ℎ
𝑓(0) − 8𝑓(0.25) + 8𝑓(0.75) − 𝑓(1)
𝑓´(0.5) =
12(0.25)
1.2 − 8(1.103515625) + 8(0.636328125) − 0.2
𝑓´(0.5) = = −𝟎. 𝟗𝟏𝟐𝟓
12(0.25)

𝑃𝑟𝑜𝑔𝑟𝑒𝑠𝑖𝑣𝑎𝑠:

## −3𝑓(𝑥0 ) + 4𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 + 2ℎ)

𝑓´(𝑥0 ) =
2ℎ
−3𝑓(0.5) + 4𝑓(0.75) − 𝑓(1)
𝑓´(0.5) =
2(0.25)
−3(0.925) + 4(0.636328125) − 0.2
𝑓´(0.5) = = −𝟎. 𝟖𝟓𝟗𝟑𝟕𝟓
2(0.25)
user
The text passage contains a series of mathematical questions and their corresponding answers, all related to calculating the first derivative of a given function using different finite difference methods.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS