For what values of K is the function f(x) = x | vedclass

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(A) For a function $f(x)$ to be strictly increasing for all $x \in \mathbb{R}$,its derivative $f'(x)$ must be greater than $0$ for all $x \in \mathbb{

Vedclass · Accepted Answer

$K > \frac{3}{2}$

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(A) For a function $f(x)$ to be strictly increasing for all $x \in \mathbb{R}$,its derivative $f'(x)$ must be greater than $0$ for all $x \in \mathbb{R}$.Given $f(x) = x^3 + 6x^2 + (9 + 2K)x + 1$,we find the derivative:$f'(x) = 3x^2 + 12x + (9 + 2K)$.For $f'(x) > 0$ for all $x$,the quadratic expression $ax^2 + bx + c$ must have $a > 0$ and the discriminant $D Here,$a = 3 > 0$,which is satisfied.The discriminant $D = b^2 - 4ac $D = (12)^2 - 4(3)(9 + 2K) $144 - 12(9 + 2K) Divide by $12$:$12 - (9 + 2K) $12 - 9 - 2K $3 - 2K $3 $K > \frac{3}{2}$.

For what values of $K$ is the function $f(x) = x^3 + 6x^2 + (9 + 2K)x + 1$ strictly increasing for all $x \in \mathbb{R}$?

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