If f(x) = x^5 - 20x^3 + 240x,then f(x) satisf | vedclass

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

Question

(C) Given the function $f(x) = x^5 - 20x^3 + 240x$. To determine the monotonicity,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(x^5 - 20x^3 +

Vedclass · Accepted Answer

It is monotonically increasing everywhere

Vedclass · Answer

(C) Given the function $f(x) = x^5 - 20x^3 + 240x$. To determine the monotonicity,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(x^5 - 20x^3 + 240x) = 5x^4 - 60x^2 + 240$. We can factor out $5$: $f'(x) = 5(x^4 - 12x^2 + 48)$. Now,we complete the square for the expression inside the parenthesis: $x^4 - 12x^2 + 48 = (x^2 - 6)^2 - 36 + 48 = (x^2 - 6)^2 + 12$. Thus,$f'(x) = 5[(x^2 - 6)^2 + 12]$. Since $(x^2 - 6)^2 \ge 0$ for all $x \in \mathbb{R}$,it follows that $(x^2 - 6)^2 + 12 \ge 12$. Therefore,$f'(x) \ge 5 	imes 12 = 60 > 0$ for all $x \in \mathbb{R}$. Since $f'(x) > 0$ for all $x$,the function $f(x)$ is monotonically increasing everywhere.

If $f(x) = x^5 - 20x^3 + 240x$,then $f(x)$ satisfies which of the following?

Similar Questions

$f(x) = x^2 - 6x + 10$ is an increasing function in the . . . . . . interval.

What type of function is $y = x^4$?

The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

If $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$ decreases for all values of $x$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module