Question

It si 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

$f\left(x\right)=x^4-62x^2+ax+9$

$f^{\prime }\left(x\right)=4x^3-124x+a$

f attains its maximum value on the interval [0,2] at x=1.

Therefore,

$f^{\prime }\left(1\right)=0$

$4-124+a=0$

$a=120$

Hence, the value of a is 120.