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

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

{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 $x = 0$