# Limits Continuity Differentiability And Differentiation Set-6

**Example – 15**

If \(f(xy) = f(x) + f(y)\) for all *x*, *y* > 0 and \(f(x)\) is differentiable at *x* = 1, then prove that \(f(x)\) is differentiable for all *x* > 0

**Solution: ** Substituting \(x = y = 1,\) we get \(f(1) = 0\)

Since \(f(x)\) is differentiable at \(x = 1,f'(1)\) exists

\(\begin{align}f'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{h}\end{align}\)

\( \begin{align}= \mathop {\lim }\limits_{h \to 0} \frac{{f(1 + h)}}{h}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots (i)\end{align}\)

Now, \(\begin{align}f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}\end{align}\)

To evaluate \(f'(x),\) we have to somehow manipulate its expression so that we are able to use the expression for \(f'(1)\) we evaluated in (i)

We do this as follows:

\(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{f\left\{ {x\left( {1 + \frac{h}{x}} \right)} \right\} - f(x)}}{h}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{\left\{ {f(x) + f\left( {1 + \frac{h}{x}} \right)} \right\} - f(x)}}{h} \left\{ \begin{align}&Using\;the\;relation\\&given\;in\;the\;question \end{align} \right\}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + \frac{h}{x}} \right)}}{{x.\frac{h}{x}}} \left\{ {Introduction\,of\,x\,in\,the\,denominator} \right\}\)

\( = \mathop {\lim }\limits_{\theta \to 0} \frac{{f(1 + \theta )}}{{x \cdot \theta }} \left\{ {\theta = \frac{h}{x}} \right\}\)

\( = \frac{{f'(1)}}{x} \left\{ {Using\, (i)} \right\}\)

Therefore, \(f'(x)\) has a finite value for all *x* > 0

\( \Rightarrow \) \(f(x)\) is differentiable everywhere.

**Example - 16** \(f(x)={{x}^{2}}-2|x|~\text{ }and~\text{ }g(x)=\left\{ \begin{align} & Min\{f(t)\,\,\,:-2\le t\le x,\,\,\,\,-2\le x<0\} \\ & Max\{f(t):\,\,\,\,\,0\le t\le x,\,\,\,\,\,\,\,\,0\le x\le 3\} \\ \end{align} \right\}\) draw the graph of \(g(x)\) and discuss its continuity and differentiability.

**Solution: ** The graph of *f*(*x*) is sketched below (in the relevant domain):

Now we must understand what the definition of *g*(*x*) means.

Consider the upper definition of *g*(*x*):

\(g(x) = Min\{ f(t): - 2 \le t \le x,\,\,\,\, - 2 \le x < 0\} \)

To evaluate \(g(x)\) at any *x*, we scan the entire interval from –2 to *x*, (this is what the variable *t* is for), select that value of *f*(*t*) which is minimum in this interval; this minimum value of *f*(*t*) becomes the value of the function at *x*.

The figure below illustrates this graphically for four different values of *x* (we are considering the interval \( - 2 \le x < 0,\) the upper definition of *g*(*x*)):

Notice that where *x* crosses –1 (when –1 < *x* < 0), the minimum value of the function in the interval [–2, *x*] becomes fixed at *t* = –1. When –2 < *x* < –1, the minimum value of the function in the interval [–2, *x*] is at *t* = *x*.

So, how do we draw the graph of *g *(*x*)? In [–2, –1], the graph of *g *(*x*) will the same as *f *(*x*) (because the minimum value of *f *(*x*) is at *t* = *x* itself, as described above). In [–1, 0], the minimum value becomes fixed at *t* = –1, equal to –1, so that in this interval the graph of * g*(

*x*) is constant;

*g*(

*x*) = 1 for \(x \in \left[ { - 1,0} \right]\) . The graph of

*g*(

*x*) for \(x \in \left[ { - 2,0} \right]\) is sketched below;

For \(0 \le x \le 3,\) the definition is \(g\left( x \right) = {\rm{Max}}\left\{ {f\left( t \right);0 \le t \le x,0 \le x \le 3} \right\}\) . The figure below illustrates how to obtain *g*(*x*) in this case for four different values of *x*:

Notice then untill x lies in the interval [0, 2], the maximum value of f(x) in the interval [0, x] is at t = 0, equal to 0.

As soon as x becomes greater than 2, the maximum value of f (x) in the interval [0, x] is now at t = x.

The graph of g(x) for the interval [0, 3] is sketched below:

The overall graph for g (x) is therefore:

We see that g (x) is discontinuous at x = 0 and non-differentiable at x = 0, 2.

Try to draw the following graphs on your own:

**(i)** \(f\left( x \right) = \max \left\{ {2x - {x^2};0 \le t \le x,0 \le x \le 2} \right\}\)

**(ii)** \(f\left( x \right) = \min \left\{ {\left| {{x^2} - 1} \right|; - 2 \le t \le x, - 2 \le x \le 2} \right\}\)