Question

(a) If \(f\left( x \right) = x + \frac{{\bf{1}}}{x}\), find \(f'\left( x \right)\).

(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of f and \(f'\).

Short answer

a) The derivative is \(f'\left( x \right) = 1 - \frac{1}{{{x^2}}}\).

b) The graph is shown below:

For the graphs, the points where \(f'\left( x \right)\) is 0, the tangent to the curve of \(f\left( x \right)\) is horizontal. The points where \(f'\left( x \right)\) is positive, the slope of \(f\left( x \right)\) is also positive. Both \(f\left( x \right)\) and \(f'\left( x \right)\) are discontinuous at \(x = 0\).

Step-by-step solution

Find the derivative of the function \(f\left( x \right) = x + \frac{{\bf{1}}}{x}\)

The derivative \(f'\left( x \right)\) can be calculated using the formula \(f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\):

\(\begin{aligned}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\left( {x + h} \right) + \frac{1}{{\left( {x + h} \right)}}} \right) - \left( {x + \frac{1}{x}} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{{{\left( {x + h} \right)}^2} + 1}}{{x + h}}} \right) - \left( {\frac{{{x^2} + 1}}{x}} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{x{{\left( {x + h} \right)}^2} + x - \left( {x + h} \right)\left( {{x^2} + 1} \right)}}{{xh\left( {x + h} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^3} + 2h{x^2} + x{h^2} + x} \right) - \left( {{x^3} + x + h{x^2} + h} \right)}}{{xh\left( {x + h} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{h{x^2} + x{h^2} - h}}{{xh\left( {x + h} \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} + xh - 1}}{{\left( {x + h} \right)x}}\\ & = 1 - \frac{1}{{{x^2}}}\end{aligned}\)

So, the value of \(f'\left( x \right)\) is \(1 - \frac{1}{{{x^2}}}\).

Sketch the graph of f and \(f'\)

Use the following commands to find the graph of \(f'\left( x \right)\).

1. In the graphing calculator, select “STAT PLOT” and enter the equation \(X + \frac{1}{X}\) and \(1 - \frac{1}{{{X^2}}}\) in the \({Y_1}\), and \({Y_2}\) tab respectively.
2. Enter the graph button in the graphing calculator.

On comparing the two graphs the points where \(f'\left( x \right)\) is 0, the tangent to the curve of \(f\left( x \right)\) is horizontal. The points where \(f'\left( x \right)\) is positive the slope of \(f\left( x \right)\) is also positive.

Both \(f\left( x \right)\) and \(f'\left( x \right)\) are discontinuous at \(x = 0\).