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Let f(x) be a monic polynomial of degree 4.  If f(1) = 1, f(2) = 2, f(3) = 3, and f(4) = 4, then what is f(6)?

 Nov 21, 2019
 #1
avatar+23542 
+3

Let \(f(x)\) be a monic polynomial of degree \(4\).  If \(f(1) = 1\), \(f(2) = 2\), \(f(3) = 3\), and \(f(4) = 4\), then what is \(f(6)\)?

 

\(\begin{array}{|lrcll|} \hline & \mathbf{f(x)} &=& \mathbf{x^4+ax^3+bx^2+cx+d} \\\\ f(1)=1: & 1&=& 1^4+a*1^3+b*1^2+c*1+d \\ & \mathbf{a+b+c+d} &=& \mathbf{0} \\\\ f(2)=2: & 2&=& 2^4+a*2^3+b*2^2+c*2+d \\ & \mathbf{8a+4b+2c+d} &=& \mathbf{-14} \\\\ f(3)=3: & 3&=& 3^4+a*3^3+b*3^2+c*3+d \\ & \mathbf{27a+9b+3c+d} &=& \mathbf{-78} \\\\ f(4)=4: & 4&=& 4^4+a*4^3+b*4^2+c*4+d \\ & \mathbf{64a+16b+4c+d} &=& \mathbf{-252} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{a+b+c+d} &=& \mathbf{0} \\ (2) & \mathbf{8a+4b+2c+d} &=& \mathbf{-14} \\ (3) & \mathbf{27a+9b+3c+d} &=& \mathbf{-78} \\ (4) & \mathbf{64a+16b+4c+d} &=& \mathbf{-252} \\ \hline (2)-(1):& 7a+3b+c &=& -14 \quad &| \quad [1] \\ (3)-(2):& 19a+5b+c &=& -64 \quad &| \quad [2] \\ (4)-(3):& 37a+7b+c &=& -174 \quad &| \quad [3] \\ \hline [2]-[1]:& 12a+2b &=& -50 \quad &| \quad \{1\} \\ [3]-[2]:& 18a+2b &=& -110 \quad &| \quad \{2\} \\ \hline \{2\}-\{1\}:& 6a &=& -60 \\ & \mathbf{a} &=& \mathbf{-10} \\\\ \{1\}& 2b &=& -50 -12a \\ & 2b &=& -50 -12*\left(-10\right) \\ & \mathbf{b} &=& \mathbf{35} \\\\ [1] & c&=& -14-7a-3b \\ & c&=& -14-7*\left(-10\right)-3*35 \\ & \mathbf{c} &=& \mathbf{-49} \\\\ (1) & d &=& -a-b-c \\ & d &=& -\left(-10\right)-35-\left(-49\right) \\ & \mathbf{d} &=& \mathbf{24} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{x^4+ax^3+bx^2+cx+d} \quad | \quad a=-10,\ b=35,\ c=-49,\ d=24 \\\\ \mathbf{f(x)} &=& \mathbf{x^4-10x^3+35x^2-49x+24} \\\\ f(6) &=& 6^4-10*6^3+35*6^2-49*6+24 \\\\ \mathbf{f(6)} &=& \mathbf{126} \\ \hline \end{array}\)

 

laugh

 Nov 21, 2019
edited by heureka  Nov 22, 2019
 #2
avatar+67 
+2

Here's a different method, Newton Interpolation, sort of.

 

Let

 \(\displaystyle f(x) = a_{0}+a_{1}(x-1)+a_{2}(x-1)(x-2)+a_{3}(x-1)(x-2)(x-3)+(x-1)(x-2)(x-3)(x-4)\)

 

(The coefficient for the final term will be 1 since the polynomial is monic).

 

Sustituting x = 1, 2, 3 ,4, gets us

 

\(\displaystyle f(1)=1=a_{0}, \qquad\text{ so }\quad a_{0}=1,\\ f(2)=2=a_{0}+a_{1},\qquad \text{ so }\quad a_{1}=1,\\ f(3)=3=a_{0}+2a_{1}+2a_{2}\qquad \text{ so }\quad a_{2}=0,\\ f(4)=4=a_{0}+3a_{1}+6a_{2}+6a_{3} \qquad\text{ so }\quad a_{3}=0.\)

 

So,

\(\displaystyle f(x)=1+(x-1)+(x-1)(x-2)(x-3)(x-4)\\ \text{and therefore}\\ f(6)=1 + 5+5*4*3*2=126.\)

.
 Nov 21, 2019
 #3
avatar+105971 
+1

Thanks Heureka and Tiggsy.   laugh

Melody  Nov 21, 2019

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