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.

 

 

Download practice questions along with solutions for FREE:
Applications of Derivatives
grade 11 | Questions Set 1
Applications of Derivatives
grade 11 | Answers Set 1
Applications of Derivatives
grade 11 | Questions Set 2
Applications of Derivatives
grade 11 | Answers Set 2