Question

Which of the following function(s) has/have point of inflection?

• A. $f\left(x\right)=x^{6/7}$
• B. $f\left(x\right)=x^6$
• C. $f\left(x\right)=\cos x+2x$
• D. $f\left(x\right)=x|x|$

Answer 1

Answer: (C), (D)

$f\left(x\right)=x|x|$

$f\left(x\right)=x^{6/7}\Rightarrow f^{\prime }\left(x\right)=\frac{6}{7}{x}^{-1/7}$
$\Rightarrow f^{\prime \prime }\left(x\right)=-\frac{6}{7}\left(\frac{1}{7}\right)x^{-8/7}$
Here, $f^{\prime \prime }\left(x\right)$ does not change sign. Hence, it has no point of inflection.

For $f\left(x\right)=x^6,f^{\prime \prime }\left(x\right)=30x^4$ but $f^{\prime \prime }\left(x\right)$ does not change sign in the neighbourhood of $x=0$
$f\left(x\right)=\cos x+2x$
or $f^{\prime \prime }\left(x\right)=-\cos x$
or $f^{\prime \prime }\left(0\right)=0$ for $x=(2n+1)\frac{\pi }{2},n\in Z$
Also, $f^{\prime \prime }\left(x\right)$ changes sign in the neighbourhood of $(2n+1)\frac{\pi }{2}$. Hence, function has infinite point of inflection.
$f\left(x\right)=x|x|=\{\begin{array}{cc}-x^2,& x<0\\ x^2,& x\ge 0\end{array}$
or $f^{\prime \prime }\left(x\right)=\{\begin{array}{cc}2,& x<0\\ 2,& x>0\end{array}$
Here, $f^{\prime \prime }\left(x\right)$ changes sign in the neighbourhood of $x=0$. Hence, has point of inflection.