Question

(a) Graph the function \({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\frac{{\bf{1}}}{{{\bf{1 + }}{{\bf{e}}^{\frac{{\bf{1}}}{{\bf{x}}}}}}}\).

(b) Explain the shape of the graph by computing the limits of \({\bf{f}}\left( {\bf{x}} \right)\) as \({\bf{x}}\) approaches \(\infty , - \infty ,{{\bf{0}}^{\bf{ + }}}{\bf{and}}\;{{\bf{0}}^{\bf{ - }}}\).

(c) Use the graph of \(f\) to estimate the coordinates of the inflection points.

(d) Use your CAS to compute and graph \({\bf{f''}}\).

(e) Use the graph in part (d) to estimate the inflection points more accurately.

Short answer

1. The graph of the function is obtained.
2. The graph of the function is obtained.
3. The inflection point occurs at \(x \approx \pm 0.4\).
4. The graph of the function is obtained.
5. The inflection point occurs at \(x \approx \pm 0.42\).

Step-by-step solution

(a) Graph of the given function.

The graph for the function \(f\left( x \right) = \frac{1}{{\left( {1 + {e^{\frac{1}{x}}}} \right)}}\) is shown below:

Adequate viewing rectangle is \(\left( { - 2.5,2.5} \right) \times \left( { - 1,2} \right)\).

(b) The shape of the graph limit \(f\left( x \right)\).

The limit function \(f\left( x \right)\), \(x\) approaches \(\infty \),

\(\begin{aligned}{c}\mathop {{\rm{lim}}}\limits_{x \to \infty } f\left( x \right) &= \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{1 + {e^{\frac{1}{x}}}}}\\ = \frac{1}{{1 + {e^0}}}\\ &= \frac{1}{{1 + 1}}\\ = \frac{1}{2}\end{aligned}\)

The limit function \(f\left( x \right)\), \(x\) approaches \( - \infty \),

\(\begin{aligned}{c}\mathop {{\rm{lim}}}\limits_{x \to - \infty } f\left( x \right) &= \mathop {{\rm{lim}}}\limits_{x \to - \infty } \frac{1}{{1 + {e^{\frac{1}{x}}}}}\\ = \frac{1}{{1 + {e^0}}}\\ &= \frac{1}{{1 + 1}}\\ &= \frac{1}{2}\end{aligned}\)

The limit function \(f\left( x \right)\), \(x\) approaches \({0^ - }\),

\(\begin{aligned}{c}\mathop {{\rm{lim}}}\limits_{x \to {0^ - }} f\left( x \right) &= \mathop {{\rm{lim}}}\limits_{x \to {0^ - }} \frac{1}{{1 + {e^{\frac{1}{x}}}}}\\ &= \frac{1}{{1 + {e^{ - \infty }}}}\\ &= \frac{1}{{1 + 0}}\\ &= 1\end{aligned}\)

The limit function \(f\left( x \right)\), \(x\) approaches \({0^ + }\),

\(\begin{aligned}{c}\mathop {{\rm{lim}}}\limits_{x \to {0^ + }} f\left( x \right) &= \mathop {{\rm{lim}}}\limits_{x \to {0^ + }} \frac{1}{{1 + {e^{\frac{1}{x}}}}}\\ &= \frac{1}{{1 + {e^\infty }}}\\ &= \frac{1}{{1 + \infty }}\\ &= 0\end{aligned}\)

Thus, the graph is below and the dotted line in graph is \(y = 0.5\).

(c) The coordinate of the inflection points.

The inflection occurs when \(f\left( x \right)\) is continuous and \(f\left( x \right)\) changes its concavity.

From the graph the inflection point occurs at \(x \approx \pm 0.4\).

Hence, the inflection point occurs at \(x \approx \pm 0.4\).

(d) Graph of double derivative of the given function.

Use calculator to compute the second derivative of the given function gives,

\(f''\left( x \right) = - \frac{{{e^{\frac{1}{x}}}\left( {2x + {e^{\frac{1}{x}}}\left( {2x - 1} \right) + 1} \right)}}{{{{\left( {1 + {e^{\frac{1}{x}}}} \right)}^3}{x^4}}}\)

The graph for \(f''\left( x \right)\) is shown below:

(e) Find inflection points more accurately.

Inflection occurs at a point if and only if, \(f\left( x \right)\) is continuous and \(f''\left( x \right)\) changes its sign.

From the graph, \(f''\left( x \right)\) changes its sign when \(x \approx - 0.42,0,0.42\).

Inflection does not occur at \(x = 0\) because \(f\left( x \right)\) is not continuous at \(x = 0\).

Hence, the inflection occurs when \(x \approx \pm 0.42\).