Problem

The equation $ y" + y' - 2y = x^2 $ is called a d…

02:04

Problem 67 Hard Difficulty

Find a second-degree polynomial & P & such that $ P(2) = 5, P'(2) = 3, $ and $ P"(2) = 2. $

Name: find a second degree polynomial p such that p2 5 p2 3 and p2 2
Uploaded: 2019-11-24
Duration: 1 min 43 s
Channel: Numerade
Description: Find a second-degree polynomial & P & such that $ P(2) = 5, P'(2) = 3, $ and $ P"(2) = 2. $

Answer

Let $P(x)=a x^{2}+b x+c$. Then $P^{\prime}(x)=2 a x+b$ and $P^{\prime \prime}(x)=2 a . P^{\prime \prime}(2)=2 \Rightarrow 2 a=2 \Rightarrow a=1$
$P^{\prime}(2)=3 \Rightarrow 2(1)(2)+b=3 \Rightarrow 4+b=3 \Rightarrow b=-1$
$P(2)=5 \Rightarrow 1(2)^{2}+(-1)(2)+c=5 \Rightarrow 2+c=5 \Rightarrow c=3 .$ So $P(x)=x^{2}-x+3$

Discussion

You must be signed in to discuss.

Top Calculus 1 / AB Educators

Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Caleb E.

Baylor University

Joseph L.

Boston College

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Video Transcript

it's clear. So when you read here so we have a P of X is equal to a X square plus B x plus c. So our first derivative, it has to be to a X plus be in our second derivative. Must to be too, eh? We're gonna plug in to what's to pay Becomes a is equal to one. So our first derivative is equal to two x plus being the first orbit of to a sequel to two comes too close. Be just equal to three. He is equal to negative one. So P f X is equal to x square minus X plus c. You have PF two is equal to two square minus two plus C, which is equal to five. C is equal to three, so you get your backs is equal to X squared minus X plus three.

Clarissa N.

Topics

Derivatives

Differentiation

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 1

Derivatives of Polynomials and Exponential Functions