The function f(x) = 1 - x^3 - x^5 is a decrea | vedclass

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

Question

(D) Given the function $f(x) = 1 - x^3 - x^5$. To determine where the function is decreasing,we find its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(1 -

Vedclass · Accepted Answer

$\forall x \in \mathbb{R}$

Vedclass · Answer

(D) Given the function $f(x) = 1 - x^3 - x^5$. To determine where the function is decreasing,we find its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(1 - x^3 - x^5) = -3x^2 - 5x^4$. We can factor out $-x^2$: $f'(x) = -x^2(3 + 5x^2)$. Since $x^2 \geq 0$ and $(3 + 5x^2) > 0$ for all $x \in \mathbb{R}$,it follows that $f'(x) = -x^2(3 + 5x^2) \leq 0$ for all $x \in \mathbb{R}$. Since $f'(x) \leq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is a decreasing function for all $x \in \mathbb{R}$.

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

Similar Questions

If $f(x) = \log(1 + x) - \frac{x}{1 + x}$, then the values of $x$ for which $f(x)$ is monotonically increasing and monotonically decreasing are respectively.....

If $f(x) = x^2 + kx + 1$ is an increasing function on the interval $[1, 2]$,what is the minimum value of $k$?

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

The function $f(x) = 2x^3 - 9x^2 + 12x + 2$ is decreasing in

The function $f(x) = \frac{\log x}{x}$ is increasing in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module