Examples on Graphing Using Derivatives Set 2

Go back to  'Applications of Derivatives'

Example - 37

Plot the graph of \(f\left( x \right) = x{e^{1/x}}\)

Solution: As far as possible, we will try to stick to the general sequence mentioned above for analysing any given function.

* The domain of \(f\left( x \right){\text{is}}\,\mathbb{R}\backslash \left\{ 0 \right\}\).

* Also, \(f\left( x \right)\) is continuous and differentiable on \(\mathbb{R}\backslash \{ 0\} \).

* Now, \(\mathop {\lim }\limits_{x \to {0^ + }} \left( {f\left( x \right)} \right) = \infty .\) Hence, x = 0 is a vertical asymptote to f (x).

  \(\mathop {\lim }\limits_{x \to {0^ - }} \left( {f\left( x \right)} \right) = 0.\)

  \(\begin{align}\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\frac{{f\left( x \right)}}{x}} \right) = \mathop {\lim }\limits_{x \to  \pm \infty } \left( {{e^{\frac{1}{x}}}} \right) = 1\end{align}\)

Also,   \(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {f\left( x \right) - x} \right) = \mathop {\lim }\limits_{x \to  \pm \infty } x\left( {{e^{\frac{1}{x}}} - 1} \right)\)

\(\begin{align}&= \mathop {\lim }\limits_{y \to 0} \frac{{{e^y} - 1}}{y}\,\,\,\,\,\,\,\left( {{\rm{where}}\,y = \frac{1}{x}} \right)\\ &= 1\end{align}\)

\( \Rightarrow \quad   y = x + {\rm{ }}1\) is an inclined asymptote to f(x).

* Quoted from I.A. Maron: Problems in Calculus

\(\begin{align} &*  \qquad f'\left( x \right) = x{e^{1/x}} \cdot \left( {\frac{{ - 1}}{{{x^2}}}} \right) + {e^{1/x}}\\&\qquad\quad\qquad = {e^{1/x}}\left( {1 - \frac{1}{x}} \right)\\&\;\;\qquad f'\left( x \right) > 0\,\,\,{\rm{for}}\,\,\,x \in \left( { - \infty ,0} \right) \cup \left( {1,\infty } \right)\\&\;\;\qquad f'\left( x \right) < 0\,\,\,{\rm{for}}\,\,\,x \in \left( {0,1} \right)\\&\;\;\qquad f'\left( x \right) = 0\,\;{\rm{for}}\,\;x = 1;\,\ \text{ a  local  minimum  point;} f(1) = e \end{align}\)

Example - 38

Sketch the graph of \(f\left( x \right) = {x^6} - 3{x^4} + 3{x^2} - 5\).

Solution:

 *          The domain is obviously \(\mathbb{R}\)

 *          (x) is an even function

 *         Since (x) is a polynomial function, it is continuous and differentiable on \(\mathbb{R}\).

 *         It is obvious that there are no asymptotes to (x)

\(\begin{align}&*\qquad f'\left( x \right) = 6{x^5} - 12{x^3} + 6x\\ & \qquad\qquad\quad\; = 6x\left( {{x^4} - 2{x^2} + 1} \right)\\ &\qquad\qquad\quad\;  = 6x{\left( {{x^2} - 1} \right)^2}\\&\qquad \;\;\;f'\left( x \right) = 0\,\,{\rm{for}}\,\,x = 0, \pm 1\\  & \qquad\qquad\quad\;= \left( {{x^2} - 1} \right)\left\{ {6\left( {{x^2} - 1} \right) + 24{x^2}} \right\}\\& \qquad\qquad\quad\;  = \left( {{x^2} - 1} \right)\left( {30{x^2} - 6} \right) \ldots  \ldots \ldots \ldots(1)\\ &\qquad\qquad\quad\;  = 6\left( {5{x^4} - 6{x^2} + 1} \right) \\&\qquad\;\;\;\ f\left( 0 \right) = 6,\,\,f\left( { \pm 1} \right) = 0  \end{align}\)

\( \Rightarrow \quad   x = {\rm{ }}0\) is a point of local minimum and \(x =  \pm 1\) are points of inflexion (verify that f''(x) does not change sign as x crosses \( \pm 1\)).

Now, \(f'\left( x \right) > 0\,\,{\rm{if}}\,\,x > 0\) and \(f'\left( x \right) < 0\) if \(x < 0\). Therefore, \(f\left( x \right)\) decreases on \(\left( { - \infty ,0} \right)\) and increases on \(\left( {0 - \infty } \right)\)

There is one more important fact we must take into account. \(f''\left( x \right)\) has roots \( \pm 1\) and additionally, \(\begin{align} \pm \frac{1}{{\sqrt 5 }}\end{align}\))  (from (i)).

Therefore, at these four points the convexity of the graph changes:

\(\begin{align}\Rightarrow  \quad f\left( x \right) > 0\,\,\,\forall \,x \in \left( { - \infty , - 1} \right) \cup \left( {\frac{{ - 1}}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right)\end{align}\)\( \cup \left( {1,\infty } \right)\) so that \(f\left( x \right)\) is concave upwards in these intervals

\(\begin{align} \Rightarrow  \quad f\left( x \right) < 0\,\,\,\forall \,\,x \in \left( { - 1,\frac{{ - 1}}{{\sqrt 5 }}} \right) \cup \left( {\frac{1}{{\sqrt 5 }},1} \right)\end{align}\) so that f(x) is concave downwards in these intervals.

*    \(f\left( 0 \right) =  - 5,f\left( { \pm 1} \right) =  - 4,\,\,f\left( { \pm 2} \right) = 23\)

Therefore one root each of f(x) lies in (–2, –1) and (1, 2)

This information is sufficient to accurately draw the graph of the given function.