In which interval is the given function f(x) | vedclass

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

Question

(B) Given function: $f(x) = 2x^3 - 15x^2 + 36x + 1$. Find the derivative: $f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) = 6x^2 - 30x + 36$. Factor the

Vedclass · Accepted Answer

$(2, 3)$

Vedclass · Answer

(B) Given function: $f(x) = 2x^3 - 15x^2 + 36x + 1$. Find the derivative: $f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) = 6x^2 - 30x + 36$. Factor the derivative: $f'(x) = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$. For the function to be monotonically decreasing,we must have $f'(x) So,$6(x - 2)(x - 3) This inequality holds when $x$ lies between the roots $2$ and $3$. Therefore,the interval is $(2, 3)$.

In which interval is the given function $f(x) = 2x^3 - 15x^2 + 36x + 1$ monotonically decreasing?

Similar Questions

The complete set of values of $m$ for which the function $f(x) = e^{\sin x} + 2m\sin x + 1$ is increasing for all $x \in \left( 0, \frac{\pi}{2} \right)$ is:

The interval in which $y = \ln(\ln(x)), x > 1$ is decreasing is

$y = x^2 (x - 3)^2$ increases for all values of $x$ lying in the interval :

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

The function $f(x) = \cot^{-1} x + x$ is increasing in the interval.

Vedclass Test Series

Exam Paper Generator

Online Exam Module