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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 $x \in \mathbb{R}$?

Similar Questions

If $f(x)=x^3+b x^2+c x+d$ and $0 < b^2 < c$,then in $(-\infty, \infty)$

Find the intervals in which the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing or strictly decreasing.

The function $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$ is

If $f(x) = x^3 - 6x^2 + 9x + 3$ is a decreasing function,then in which interval does $x$ lie?

The function $y = 6 - 9x - x^2$ is strictly increasing on the interval . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module