For some function \(f(x)=ax^3+bx^2+cx+d\), where \(a\), \(b\), \(c\), and \(d\) are integers, we know the following information:
the \(y\)-intercept is \(5\),
\(f(2)=-3\),
\(f(4)\) is greater than \(40\) but less than \(50\), and
\(f(6)\) is greater than \(240\) but less than \(250\).
Determine the value of \(f(7)\).
Since the \(y\)-intercept is \(5\), it follows that \(f(0)=5\). Thus, \[\begin{align} a(0)^3+b(0)^2+c(0)+d &=5\\ d &=5 \end{align}\] We can now write the function as \(f(x)=ax^3+bx^2+cx+5\).
Since \(f(2)=-3\), \[\begin{align} a(2)^3+b(2)^2+c(2)+5 &= -3\\ 8a+4b+2c+5 &= -3\\ 8a+4b+2c &= -8\\ 4a+2b+c &= -4\\ c &= -4a-2b-4 \tag{1} \end{align}\]
Next we consider \(f(4)\). \[\begin{align} f(4) &= a(4)^3+b(4)^2+c(4)+5\\ &= 64a+16b+4c+5\\ &= 64a+16b+4(-4a-2b-4)+5 \tag{using equation $(1)$}\\ &= 64a + 16b - 16a - 8b - 16 + 5\\ &= 48a + 8b - 11 \end{align}\] Since \(f(4)>40\), it follows that \(48a + 8b - 11>40\) or \(48a+8b>51\). Dividing the inequality by \(8\) gives \(6a+b>6.375\). Similarly, since \(f(4)<50\), it follows that \(48a + 8b - 11<50\) or \(48a+8b<61\). Dividing the inequality by \(8\) gives \(6a+b<7.625\). Since \(a\) and \(b\) are integers it follows that \(6a+b\) is an integer. Thus, since \(6a+b>6.375\) and \(6a+b<7.625\), we can conclude that \(6a+b=7\).
Next we consider \(f(6)\). \[\begin{align} f(6) &= a(6)^3+b(6)^2+c(6)+5\\ &= 216a+36b+6c+5\\ &= 216a+36b+6(-4a-2b-4)+5 \tag{using equation $(1)$}\\ &= 216a + 36b -24a - 12b - 24 + 5\\ &= 192a + 24b - 19 \end{align}\] Since \(f(6)>240\), it follows that \(192a + 24b - 19>240\), or \(192a+24b>259\). Dividing the inequality by \(24\) gives \(8a+b>10 \frac{19}{24}\). Similarly, since \(f(6)<250\), it follows that \(192a + 24b - 19<250\), or \(192a+24b<269\). Dividing the inequality by \(24\) gives \(8a+b<11 \frac{5}{24}\). Since \(a\) and \(b\) are integers it follows that \(8a+b\) is an integer. Thus, since \(8a+b>10 \frac{19}{24}\) and \(8a+b<11 \frac{5}{24}\), we can conclude that \(8a+b=11\).
We now have the following system of equations. \[\begin{align} 6a+b&=7 \tag{2}\\ 8a+b&=11 \tag{3} \end{align}\] By subtracting equation \((2)\) from equation \((3)\), we obtain \(2a=4\), or \(a=2\). Substituting \(a=2\) in equation \((2)\) gives \(6(2)+b=7\), and thus \(b=-5\).
Substituting \(a=2\) and \(b=-5\) in equation \((1)\) gives: \[\begin{align} c &= -4a-2b-4\\ &= -4(2)-2(-5)-4\\ &= -8+10-4=-2 \end{align}\] We can now write the function as \(f(x)=2x^3-5x^2-2x+5\).
Finally we can determine \(f(7)\). \[\begin{align} f(7) &= 2(7)^3-5(7)^2-2(7)+5\\ &= 2(343)-5(49)-14+5\\ &= 686-245-9= 432 \end{align}\] Therefore, \(f(7)=432\).
Note:
We could have written the third bullet point as \(40<f(4)<50\) and solved the entire inequality at once instead of dealing with the inequality symbols one at a time. While this may be unfamiliar to students, it’s a helpful way to solve inequalities. This would have looked as follows. \[\begin{align} {3} 40&<f(4)&&<50\\ 40&<48a + 8b - 11&&<50\\ 51&<48a+8b&&<61\\ 6.375&<6a+b&&<7.625 \end{align}\] From here, we can conclude that since \(6a+b\) is an integer, then we must have \(6a+b=7\). We could have then used a similar approach to solve the inequalities in \(f(6)\).