If the function f(x) = x^4 - 62x^2 + ax + 9 i | vedclass

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(A) Given the function $f(x) = x^4 - 62x^2 + ax + 9$.For a function to have a local maximum or minimum at a point $x = c$,the first derivative $f'(x)$

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Given the function $f(x) = x^4 - 62x^2 + ax + 9$.For a function to have a local maximum or minimum at a point $x = c$,the first derivative $f'(x)$ must be zero at that point.First,find the derivative of $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9) = 4x^3 - 124x + a$.Since the function has a maximum at $x = 1$,we set $f'(1) = 0$:$4(1)^3 - 124(1) + a = 0$.Simplifying the equation:$4 - 124 + a = 0$.$-120 + a = 0$.$a = 120$.Thus,the value of $a$ is $120$.

If the function $f(x) = x^4 - 62x^2 + ax + 9$ is maximum at $x = 1$,then the value of $a$ is

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