Question

It is given that at x=1, the function $x^4-62x^2+ax+9$ attains its maximum value, on the interval [0,2]. Find the value of a.

Answer 1

Let $f\left(x\right)=x^4-62x^2+ax+9\Rightarrow f^{\prime }\left(x\right)=4x^3-124x+a$
It is given that function f attains its maximum value on the interval [0,2] at x=1
$\begin{array}{cc}\therefore & f^{\prime }\left(1\right)=0\\ \Rightarrow & 4\times 1^3-124\times 1+a=0\\ \Rightarrow & 4-124+a=0\Rightarrow a=120\end{array}$
Hence, the value of a is 120.