The function \(f(x)=x^5-4x^4+ax^3-x^2+bx-8\) has a value of \(128\) when \(x=4\). Determine the value of the function when \(x=-4\).
We know that the function has a value of \(128\) when \(x=4\). That is, \(f(4)=128\). Then, \[\begin{aligned} f(4) &= 128\\ (4)^5-4(4)^4+a(4)^3-(4)^2+b(4)-8 &= 128\\ 1024-1024+64a-16+4b-8 &= 128\\ 64a+4b &= 152 \end{aligned}\] We want to determine the value of the function when \(x=-4\). That is, we want to determine \(f(-4)\). Then, \[\begin{aligned} f(-4) &= (-4)^5-4(-4)^4+a(-4)^3-(-4)^2+b(-4)-8\\ &= -1024-1024-64a-16-4b-8\\ &= -64a-4b-2072 \end{aligned}\] However, we know from earlier that \(64a+4b = 152\). Then, \[f(-4) = -(64a+4b)-2072=-152-2072 = -2224\] Therefore, the value of the function is \(-2224\) when \(x=-4\).
Note that we are not given enough information to determine the values of \(a\) and \(b\), however enough information is given to determine the value of \(f(-4)\).