(A) 12

(B) 8

(C) 0

(D) -2

Elianna Lawrence
2022-07-26
Answered

If $f(x)={x}^{3}+A{x}^{2}+Bx-3$ and if f(1)=4 and f(-1)=-6, what is the value of 2A+B?

(A) 12

(B) 8

(C) 0

(D) -2

(A) 12

(B) 8

(C) 0

(D) -2

You can still ask an expert for help

Marisa Colon

Answered 2022-07-27
Author has **18** answers

$f(x)={x}^{3}+A{x}^{2}+Bx-3$ and if f(1)=4 and f(-1)=-6, what is the value of 2A+B?

$f(1)=4=1+A+B-3\Rightarrow A+B=6$.....(I)

$f(-1)=-6=-1+A-B-3\Rightarrow A-B=-2$......(II)

2A =4

$A=2\Rightarrow A+B=6\Rightarrow B=4$

Hence 2A+B= 4+4=8

$f(1)=4=1+A+B-3\Rightarrow A+B=6$.....(I)

$f(-1)=-6=-1+A-B-3\Rightarrow A-B=-2$......(II)

2A =4

$A=2\Rightarrow A+B=6\Rightarrow B=4$

Hence 2A+B= 4+4=8

Shannon Andrews

Answered 2022-07-28
Author has **5** answers

We have that f(1)=4 and that f(-1)=6. This gives us:

$f(1)=4(1{)}^{3}+A({1}^{2})+B(1)-3=A+B-2$

$f(-1)=-6=(-1{)}^{3}+A(-{1}^{2})+B(-1)-3=A-B-4$

So, we have:

4=A+B-2....(1)

-6=A-B-4.....(2)

Adding (1) and (2) gives us:

4+(-6)=(A+B-2)+(A-B-4)

-2=2A-6=> 2A=4

Now, we subtract both of them and get:

4-(-6)=(A+B-2)-(A-B-4)

10=2B+2 => B=4

Therefore, 2A+B = 4+4 = 8 , which is option (B)

$f(1)=4(1{)}^{3}+A({1}^{2})+B(1)-3=A+B-2$

$f(-1)=-6=(-1{)}^{3}+A(-{1}^{2})+B(-1)-3=A-B-4$

So, we have:

4=A+B-2....(1)

-6=A-B-4.....(2)

Adding (1) and (2) gives us:

4+(-6)=(A+B-2)+(A-B-4)

-2=2A-6=> 2A=4

Now, we subtract both of them and get:

4-(-6)=(A+B-2)-(A-B-4)

10=2B+2 => B=4

Therefore, 2A+B = 4+4 = 8 , which is option (B)

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