The maximum value of a such that the second d | vedclass

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(C) Let $f(x) = x^4+ax^3+\frac{3x^2}{2}+1$. First,find the first derivative: $f'(x) = 4x^3+3ax^2+3x$. Next,find the second derivative: $f''(x) = 12x^2

Vedclass · Accepted Answer

$2$

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(C) Let $f(x) = x^4+ax^3+\frac{3x^2}{2}+1$. First,find the first derivative: $f'(x) = 4x^3+3ax^2+3x$. Next,find the second derivative: $f''(x) = 12x^2+6ax+3$. We are given that $f''(x) > 0$ for all real $x$,so $12x^2+6ax+3 > 0$. Dividing by $3$,we get $4x^2+2ax+1 > 0$. For a quadratic $Ax^2+Bx+C > 0$ to hold for all $x$,the discriminant $D$ must be less than $0$ and $A > 0$. Here $A = 4 > 0$,so we require $D $D = (2a)^2 - 4(4)(1) $4a^2 - 16 $a^2 This implies $-2 The maximum value of $a$ is $2$.

The maximum value of $a$ such that the second derivative of $x^4+ax^3+\frac{3x^2}{2}+1$ is positive for all real $x$ is

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