If f(x) = kx^3 - 3x^2 - 12x + 8 is strictly d | vedclass

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Question

(A) Given the function $f(x) = kx^3 - 3x^2 - 12x + 8$.For $f(x)$ to be strictly decreasing for all $x \in R$,we must have $f'(x) < 0$ for all $x \in R

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$k < -\frac{1}{4}$

Vedclass · Answer

(A) Given the function $f(x) = kx^3 - 3x^2 - 12x + 8$.For $f(x)$ to be strictly decreasing for all $x \in R$,we must have $f'(x) Calculating the derivative: $f'(x) = 3kx^2 - 6x - 12$.For a quadratic expression $ax^2 + bx + c$ to be strictly negative for all $x$,we require $a Here,$a = 3k$,$b = -6$,and $c = -12$.Condition $1$: $3k Condition $2$: $D = b^2 - 4ac $(-6)^2 - 4(3k)(-12) $36 + 144k $144k $k Since $k < -\frac{1}{4}$ already satisfies $k < 0$,the final condition is $k < -\frac{1}{4}$.

If $f(x) = kx^3 - 3x^2 - 12x + 8$ is strictly decreasing for all $x \in R$,then:

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