Find the intervals in which the function f(x) | vedclass

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

Question

(A) Given function is $f(x) = 10 - 6x - 2x^2$. First,find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$. To find the critic

Vedclass · Accepted Answer

Strictly increasing on $(-\infty, -3/2)$ and strictly decreasing on $(-3/2, \infty)$

Vedclass · Answer

(A) Given function is $f(x) = 10 - 6x - 2x^2$. First,find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$. To find the critical points,set $f'(x) = 0$: $-6 - 4x = 0 \Rightarrow 4x = -6 \Rightarrow x = -\frac{3}{2}$. The point $x = -\frac{3}{2}$ divides the real line into two intervals: $(-\infty, -\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$. Case $1$: For $x \in (-\infty, -\frac{3}{2})$,let $x = -2$. $f'(-2) = -6 - 4(-2) = -6 + 8 = 2 > 0$. Thus,$f(x)$ is strictly increasing on $(-\infty, -\frac{3}{2})$. Case $2$: For $x \in (-\frac{3}{2}, \infty)$,let $x = 0$. $f'(0) = -6 - 4(0) = -6 Thus,$f(x)$ is strictly decreasing on $(-\frac{3}{2}, \infty)$.

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

Similar Questions

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

The function $f(x) = \tan^{-1}(\sin x + \cos x)$,$x > 0$ is always an increasing function on the interval

The function $f(x) = 1 - x^3 - x^5$ is a decreasing function for:

The value of $k$ such that the function $f(x) = \sin x - \cos x - kx + b$ is strictly decreasing for all real $x$ is

Let $f(x) = x^3 + bx^2 + cx + d$ where $0 < b^2 < c$. Then $f(x)$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module