MHB How do I use inequalities to find f(3) when d is 5?

[Ilikebugs]

View attachment 6582 I know that d is 5, but I don't know how to use the inequalities to find f(3)

 

[I like Serena]

I know that d is 5, but I don't know how to use the inequalities to find f(3)

Hey Ilikebugs,

Suppose we approximate f(4) and f(6) for now with, say, 45 respectively 245.
Could we solve it then?

 

[Ilikebugs]

I don't know how to solve that

 

[I like Serena]

I don't know how to solve that

Can we perhaps start by creating a set of equations?

 

[HallsofIvy]

You know that f(2)= 8a+ 4b+ 2c+ 5= -3 so that 4a+ 2b+ c= -4. From that, c= -4- 2b- 4a so we can write f(x)= ax^3+ bx^3- (4+ 2b+ 4a)x+ 5. So what is f(4)? What inequalities do you have for f(4)? Same for f(6).

Also, the fact that a, b, c, and d are integers is crucial.

 

[Ilikebugs]

I don't know, this is hard to do in my head. I got 8a^3+4b^2+2c+5=-3, 64a^3+16b^2+4c+5=45, and 216a^3+36b^2+6c+5=245

 

[Evgeny.Makarov]

this is hard to do in my head.

You should do it on paper instead. (Smile)

I got 8a^3+4b^2+2c+5=-3, 64a^3+16b^2+4c+5=45, and 216a^3+36b^2+6c+5=245

How did you get $a^3$ if in the original formula for $f$ the coefficient $a$ is never raised to any power, only $x$ is.

Follow HallsofIvy's advice.

 

[Ilikebugs]

8a+4b+2c+5=-3
64a+16b+4c+5=45
216a+36b+6c+5=245

What do I do now?

 

[Evgeny.Makarov]

What inequalities do you have for f(4)? Same for f(6).

Follow HallsofIvy's advice.

Is there a reason you don't want to do this?

8a+4b+2c+5=-3
64a+16b+4c+5=45
216a+36b+6c+5=245

In general, if you had four equations $f(x_i)=y_i$, $i=1,\dots,4$ with known $x_i$ and $y_i$, you could write four linear equations in this way and thus find the coefficients $a,\dots,d$. Here, however, you have one equation, four inequalities and the stipulation that the coefficients are integer. This makes it a little trickier. But I still would first write the four inequalities on $a$ and $b$.

 

[Ilikebugs]

f(0)=5
f(2)=8a+4b+2c+5=-3
40<f(4)=64a+16b+4c+5<50
240<f(6)=216a+36b+6c+5<250

Is this correct?

 

[Evgeny.Makarov]

Correct, but HallsofIvy also suggested expressing $c$ from the equation $f(2)=\ldots$ and substituting it in the inequalities.

 

[Ilikebugs]

-4- 2b- 4a

f(0)=5
f(2)=8a+4b+2(-4- 2b- 4a)+5=-3
40<f(4)=64a+16b+4(-4- 2b- 4a)+5<50
240<f(6)=216a+36b+6(-4- 2b- 4a)+5<250

f(0)=5
f(2)=8a+4b+(-8- 4b- 8a)+5=-3
40<f(4)=64a+16b+(-16- 8b- 16a)+5<50
240<f(6)=216a+36b+(-24- 12b- 24a)+5<250

What do I do now?

 

[Evgeny.Makarov]

Simplify the last two lines.

 

[Ilikebugs]

51<f(4)=48a+8b<61
259<f(6)=192a+24b<269

 

[Evgeny.Makarov]

Then I would multiply the first line by 3 to get
\begin{align}
153<&144a+24b<183\\
259<&192a+24b<269
\end{align}

Suppose in general that
\begin{align}
x&<\rlap{u}\hphantom{u+v}<y\\
x'&<u+v<y'
\end{align}
From the first line it follows that $-y<-u<-x$. Inequalities can be added: $a<b$ and $a'<b'$ imply that $a+a'<b+b'$. Adding $-y<-u<-x$ and $x'<u+v<y'$ produces
\[
x'-y<v<y'-x.
\]
Apply this fact to the first two (rather, four) inequalities in this post.

 

[Ilikebugs]

How do I do that?

Is a=2 and b=-5 correct?

 

[Evgeny.Makarov]

Is a=2 and b=-5 correct?

Yes.

How do I do that?

Try to compare, or match,
\begin{align}
153<&144a+24b<183\\
259<&192a+24b<269
\end{align}
and
\begin{align}
x&<\rlap{u}\hphantom{u+v}<y\\
x'&<u+v<y'
\end{align}
Such matching of your situation with a more general theorem assumption is how applying any theorem works in mathematics.

 

[Ilikebugs]

a=2 b=-5 c=-2 d=5

f(3)=2(3)^3+(-5)(3)^2+(-2)(3)+5
=54-45-6+5
=8

Is this right?

 

[Evgeny.Makarov]

Yes, this is correct, but you didn't show the crucial step: how to determine $a=2$ and $b=-5$.

 

[Ilikebugs]

153-144a<24b<183-144a
259<192a+24b<269

(259)-(183-144a)<192a<(269)-(153-144a)

66+144a<192a<116+144a
66<48a<116

Since a is an integer, a=2

Plugging in a into the 2 equations,

-135<24b<-105
-125<24b<-115

If b is an integer, b=-5

 

[Evgeny.Makarov]

All right. I took $u=144a+24b$. Then $u+v=192a+24b$ implies that $v=48a$. Thus $259-183=76<48a<116=269-153$, or $19<12a<29$, from where $a=2$.