Q. The function $f$ defined by $f(x) = 4x^{4} - 2x + 1$ in increasing for
A) $x < 1$
B) $x > 0$
C) $x < \frac{1}{2}$
D) $x > \frac{1}{2}$
Explanation
(d)
Given, $f(x) = 4x^{4} - 2x + 1$
$\Rightarrow f^{'}(x) = 16x^{3} - 2$
For maxima and minima, put $f^{'}(x) = 0$
$\Rightarrow 16x^{3} - 2 = 0$
$\Rightarrow x = \frac{1}{2}$
In interval $\left( \frac{1}{2},\infty \right)$,put $x$=1
$f^{'}(1) = 16 - 2 = 14$ [increasing]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.