Let $f(x) = \left\{ \begin{array}{r} \left| x^{3} + x^{2} + 3x + \sin x ight|\left( 3 + \sin\frac{1}{x} ight),\ x
eq 0 \ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0 \end{array} ight.\$, then number of points (where $f(x)$ attains its minimum value) is

Question

Let $f(x) = \left\{ \begin{array}{r} \left| x^{3} + x^{2} + 3x + \sin x ight|\left( 3 + \sin\frac{1}{x} ight),\ x 
eq 0 \ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0 \end{array} ight.\$, then number of points (where $f(x)$ attains its minimum value) is

Dr. Abubacker · Accepted Answer

(a)
$f(x) = \left\{ \begin{array}{r}
\left| x^{3} + x^{2} + 3x + \sin x ight|\left( 3 + \sin\left( \frac{1}{x} ight) ight),\ x 
eq 0 \
0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0
\end{array} ight.\$
Let g$(x) = x^{3} + x^{2} + 3x + \sin x$
$g^{'}(x) = 3x^{2} + 2x + 3 + \cos x$
$= 3\left( x^{2} + \frac{2x}{3} + 1 ight) + \cos x$
$= 3\left\{ \left( x + \frac{1}{3} ight)^{2} + \frac{8}{9} ight\} + \cos x > 0$
And $2 < 3 + \sin\left( \frac{1}{x} ight) < 4$
Hence, minimum value of $f(x)is\ 0\ at\ x = 0$
Hence, number of points=1

Explanation

Reviewed by Dr. Abubacker

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