Q. If $f(x) = x^{2} + 4x + 1,\ then\$
A) $f(x) = f( - x),$ for all$\ x$
B) $f(x) \neq 1,$ for all $x$=0
C) $f^{''}(x) > 0,$for all $x$
D) $f(x) > 1,$for $x \leq 4$
Explanation
(c)
Given, $f(x) = x^{2} + 4x + 1$
$\Rightarrow \ \ \ \ f^{'}(x) = 2x + 4$
For maxima or minima, put $f^{'}(x) = 0$
$\Rightarrow \ \ \ \ \ \ 2x + 4 = 0$
$\Rightarrow \ \ \ \ \ x = - 2$
Now, $f^{''}(x) = 2 > 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.