It is given that at x=1, the function f(x) = | vedclass

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(A) Let $f(x) = x^{4}-62x^{2}+ax+9.$First,we find the derivative of the function $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x^{4}-62x^{2}+ax+9)

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$120$

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(A) Let $f(x) = x^{4}-62x^{2}+ax+9.$First,we find the derivative of the function $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x^{4}-62x^{2}+ax+9) = 4x^{3}-124x+a.$It is given that the function $f(x)$ attains its maximum value on the interval $[0,2]$ at $x=1.$Since $x=1$ is an interior point of the interval $[0,2],$ the derivative at this point must be zero,i.e.,$f'(1) = 0.$Substituting $x=1$ into the derivative expression:$f'(1) = 4(1)^{3}-124(1)+a = 0.$$4-124+a = 0.$$-120+a = 0.$$a = 120.$Thus,the value of $a$ is $120.$

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

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