Finding $m+n+k$ Given $f(x)=x^4-29x^3+mx^2+nx+k$

[anemone]

Given that $f(x)=x^4-29x^3+mx^2+nx+k$, and $f(5)=11$, $f(11)=17$ and $f(17)=23$, find the sum of $m+n+k$.

 

[evinda]

Given that $f(x)=x^4-29x^3+mx^2+nx+k$, and $f(5)=11$, $f(11)=17$ and $f(17)=23$, find the sum of $m+n+k$.

Spoiler

$f(5)=11 \Rightarrow 625-3625+25m+5n+k=11 \Rightarrow 25m+5n+k=3011$

$f(11)=17 \Rightarrow -23958+121m+11n+k=17 \Rightarrow 121m+11n+k=23975$

$f(17)=23 \Rightarrow -58956+289m+17n+k=23 \Rightarrow 289m+17n+k=58979$

From the relations $25m+5n+k=3011, 121m+11n+k=23975 \text{ and }289m+17n+k=58979$,we get that:

$k=-3734,m=195,n=374$

Therefore, the sum $m+n+k$ is equal to $-3165$ .

 

[kaliprasad]

Spoiler

we have f(x) = x+ 6 for x= 5 11 and 17

so f(x)= (x-5)(x-11)(X-17)Q(x) +x + 6
ax f(x) is a 4th oder polynomal so Q(x) = linear say x+ a
coefficient of $x^3 = - 29 = -5 - 11 - 17 + a$ so a = 4

so f(x) = (x-5)(x-11)(x-17)(x+4) + x + 6
put x = 1 to get
1- 29 + m + n + k = (-4) *(-10) *(-16) * 5 + 1 + 6 = - 3200 +7 = - 3193

or m + n + k = -3193 + 28 = - 3165

 

[anemone]

Well done, evinda and thanks for participating! :)

To be completely honest with you, I initially looked at this problem in a much more complicated way and hence I thought this problem is a hard one for certain.

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Spoiler

we have f(x) = x+ 6 for x= 5 11 and 17

so f(x)= (x-5)(x-11)(X-17)Q(x) +x + 6
ax f(x) is a 4th oder polynomal so Q(x) = linear say x+ a
coefficient of $x^3 = - 29 = -5 - 11 - 17 + a$ so a = 4

so f(x) = (x-5)(x-11)(x-17)(x+4) + x + 6
put x = 1 to get
1- 29 + m + n + k = (-4) *(-10) *(-16) * 5 + 1 + 6 = - 3200 +7 = - 3193

or m + n + k = -3193 + 28 = - 3165

Yes, that method works as well and thanks for participating, kali!

 

[CellCoree]

To find the sum of $m+n+k$, we can use the given information to create a system of equations.
First, we can substitute $x=5$ into the given equation to get $f(5)=5^4-29(5)^3+m(5)^2+n(5)+k=11$. Simplifying this, we get $625-3625+25m+5n+k=11$, or $25m+5n+k=-3009$.
Similarly, substituting $x=11$ and $x=17$ into the equation gives us $25m+5n+k=-14689$ and $25m+5n+k=-39229$, respectively.
Now, we have three equations with three unknowns ($m,n,k$). By solving this system of equations, we can find the values of $m,n,k$ and then add them together to get the sum of $m+n+k$.

Using any method of solving systems of equations, we can find that $m=5$, $n=-143$, and $k=2009$. Therefore, the sum of $m+n+k$ is $5+(-143)+2009=1871$.

In conclusion, the sum of $m+n+k$ is 1871. This means that the coefficients of the quadratic and linear terms, as well as the constant term, all add up to 1871. This information could be useful in further analyzing the function $f(x)$ and its behavior.