Question

Given that $f(x) = \dfrac{{xg(x)}}{{\left| x \right|}}$, $g(0) = 0$ and $f(x)$ is continuous at $x = 0$. Then find the value of ${f^1}(0)$.

Answer 1

Hint: First we have to define what the terms we need to solve the problem are. $\left| x \right|$ is in division it cannot be zero, thus $x$ $ \ne 0$ [if $x$=$0$ then $f(x)$ turns to infinity] Depends on $x$ value only $g(x)$ maybe positive or negative signs.

Complete step-by-step solution:
Let us consider from given $f(x) = \dfrac{{xg(x)}}{{\left| x \right|}}$; $x \ne 0$
Now , find whether $f(x)$ is continuous or not.
If the value of $x > 0$ then $g(x)$ must be positive,
If the value of $x < 0$ then $g(x)$ must be negative,
From this information we can make
\[_{f(x) = \left\{ {\begin{array}{*{20}{c}}
  {g(x)}&{x > 0} \\
  { - g(x)}&{x < 0}
\end{array}} \right.}\]
Since from given, it is given that $g(0) = 0$
$\mathop {\lim }\limits_{x \to o} g(x) = g(0) = 0$ Simplifying $g(x)$ to find $f(x)$ to be continuous at $x = 0$
First, check supremum and infimum of $f(x)$
Let $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} - g(x) = - \mathop {\lim }\limits_{x \to 0} g(x) = 0$[$f(x) = - g(x)$at$x < 0$]
And $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} g(x) = \mathop {\lim }\limits_{x \to 0} g(x) = 0$ [ $f(x) = g(x)$at $x > 0$]
Therefore, from these two limits on supremum and infimum on the bounded function we get
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)$
Thus $f(x)$is continuous at $x = 0$ [satisfied bounded for supremum and infimum]
Hence $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$
Therefore $f(0) = 0$
Since $f(x)$ is continuous at $x = 0$, now to check if $f(x)$ is differentiable at $x = 0$ or not.
If the value of $x > 0$ then $g(x)$ must be positive,
If the value of $x < 0$ then $g(x)$ must be negative,
If the value of $x = 0$ then $g(x)$ must turns into zero and hence as follows
 $f(x) = $ {$g(x)$; $x > 0$, $0; x = 0$, -$g(x)$; $x < 0$
General derivative of $f(x)$ at $x$ is \[\mathop {\lim }\limits_{h \to {x^ - }} \dfrac{{f(x - h) - f(x)}}{{x - h}}\]
Now for left hand derivative (in infimum) of $f(x)$ at $x = 0$ is
$\mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{f( - h)}}{{ - h}}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{g( - h) - g(0)}}{{ - h}}$ since $g(0) = 0$
$ \Rightarrow - {g^1}(0) = 0$
Now right-hand derivative of $f(x)$ at $x = 0$ is
$ \Rightarrow {g^1}(0) = 0$
From left and right hand derivative limits we get ${g^1}(0) = 0$
Therefore $f(x)$ is differentiable at $x = 0$
Thus since $f(x)$ is continuous at $x = 0$ and also $f(x)$ is differentiable at $x = 0$
Hence ${f^1}(0) = 0$

Note: $f(x)$ is continuous only if its satisfied the bounded property of supremum and infimum[least upper bound and greatest lower bound]
Also ${f^1}(0)$ is the differentiation of $f$ with respect to zero implies of $x = 0$