For what values of x is the function f(x) = x | vedclass

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

Question

(C) To determine where the function $f(x) = x^9 + 3x^7 + 64$ is strictly increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^9 + 3x^7 +

Vedclass · Accepted Answer

All real values of $x$

Vedclass · Answer

(C) To determine where the function $f(x) = x^9 + 3x^7 + 64$ is strictly increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^9 + 3x^7 + 64) = 9x^8 + 21x^6$.Since $x^8 \geq 0$ and $x^6 \geq 0$ for all real $x$,it follows that $9x^8 + 21x^6 \geq 0$ for all $x \in \mathbb{R}$.Specifically,$f'(x) > 0$ for all $x 
eq 0$,and $f'(0) = 0$.Since the derivative is non-negative for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly increasing for all real values of $x$.

For what values of $x$ is the function $f(x) = x^9 + 3x^7 + 64$ strictly increasing?

Similar Questions

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

If $x$ lies in the interval $(0, \pi/2)$,then the function $f(x) = x \sin x + \cos x + \cos^2 x$ is:

Function $f(x) = |\sin x|$,$x \in \left(-\frac{\pi}{2}, 0\right)$ is . . . . . . .

The equation $x^3+x-1=0$ has

The function $f(x) = \frac{|x - 1|}{x^2}$ is monotonically decreasing in-

Vedclass Test Series

Exam Paper Generator

Online Exam Module