Question

(a) If \(f\left( x \right) = \left( {{x^3} - x} \right){e^x}\), 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 value of \(f'\left( x \right)\) is \({e^x}\left( {{x^3} + 3{x^2} - x - 1} \right)\).

(b) When \(f\) has a horizontal tangent line, the value of \(f'\) is zero. Moreover, \(f'\) is negative in the interval when \(f\) is decreasing and it is positive when \(f\) is increasing.

Step-by-step solution

The product rule of differentiation

If a function\(h\left( x \right) = f\left( x \right) \cdot g\left( x \right)\)where\(f\)and\(g\)are both differentiable, then the derivative of\(h\left( x \right)\)is as follows:

\(\begin{aligned}\frac{{dh}}{{dx}} & = \frac{d}{{dx}}\left( {f\left( x \right) \cdot g\left( x \right)} \right)\\ & = f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right)\end{aligned}\)

The derivative of the given function

Differentiate\(f\left( x \right) = \left( {{x^3} - x} \right){e^x}\)using the product rule as follows:

\(\begin{aligned}f\left( x \right) & = \left( {{x^3} - x} \right){e^x}\\f'\left( x \right) & = \left( {{x^3} - x} \right)\frac{d}{{dx}}\left( {{e^x}} \right) + {e^x}\frac{d}{{dx}}\left( {\left( {{x^3} - x} \right)} \right)\\& = \left( {{x^3} - x} \right)\left( {{e^x}} \right) + {e^x}\left( {3{x^2} - 1} \right)\\ & = {e^x}\left( {{x^3} - x + 3{x^2} - 1} \right)\\ & = {e^x}\left( {{x^3} + 3{x^2} - x - 1} \right)\end{aligned}\)

Hence, the value of \(f'\left( x \right)\) is \({e^x}\left( {{x^3} + 3{x^2} - x - 1} \right)\).

Check the answer visually

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

To check the answer visually draw the graph of the function \({f_1}\left( x \right) = \left( {{x^3} - x} \right){e^x}\)and \({f_2}\left( x \right) = {e^x}\left( {{x^3} + 3{x^2} - x - 1} \right)\) by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\left( {{x^3} - x} \right){e^x}\)in the\({Y_1}\)tab.
2. Select the “STAT PLOT” and enter the equation\({e^x}\left( {{x^3} + 3{x^2} - x - 1} \right)\)in the\({Y_2}\)tab.
3. Enter the “GRAPH” button in the graphing calculator.

Visualization of graph of the function\({f_1}\left( x \right) = \left( {{x^3} - x} \right){e^x}\)and \({f_2}\left( x \right) = {e^x}\left( {{x^3} + 3{x^2} - x - 1} \right)\) is shown below:

Notice that when \(f\) has a horizontal tangent line, the value of \(f'\) is zero. Moreover, \(f'\) is negative in the interval when \(f\) is decreasing and it is positive when \(f\) is increasing.