The function f(x) = 2x^3 - 6x + 5 is an incre | vedclass

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

Question

(C) Given the function $f(x) = 2x^3 - 6x + 5$. To find the intervals where the function is increasing,we first find the derivative $f'(x)$. $f'(x) = \

Vedclass · Accepted Answer

$x  1$

Vedclass · Answer

(C) Given the function $f(x) = 2x^3 - 6x + 5$. To find the intervals where the function is increasing,we first find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2x^3 - 6x + 5) = 6x^2 - 6$. $A$ function is increasing when $f'(x) > 0$. So,$6x^2 - 6 > 0$. Dividing by $6$,we get $x^2 - 1 > 0$,which factors as $(x - 1)(x + 1) > 0$. Using the sign scheme for the inequality $(x - 1)(x + 1) > 0$,the expression is positive when $x > 1$ or $x Thus,the function is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.

The function $f(x) = 2x^3 - 6x + 5$ is an increasing function,if

Similar Questions

For what value of $\lambda$ is the function $f(x) = \lambda x + \cos x$ strictly increasing?

Let $f(x)=e^x-x$ and $g(x)=x^2-x, \forall x \in R$. Then the set of all $x \in R$,where the function $h(x)=(fog)(x)$ is increasing is

The interval in which $y=x^{2} e^{-x}$ is increasing is

If $f(x) = \sin x - \cos x - ax + b$ is a decreasing function for all $x \in R$,then:

If $f(x) = x^3 - 6x^2 + 9x + 3$ is a monotonically decreasing function,then $x$ lies in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module