Question

Investigate the family of curves given by the equation \(f\left( x \right) = {x^4} + c{x^2} + x\). Start by determining the transitional value of c at which the number of inflection points changes. Then graph several members of the family to see what shapes are possible. There is another transitional value of c at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered.

Short answer

When\(c = 0\)there is no point of inflection, and the curve is a concave upward elsewhere.

When\(c\)increases, then the curve just becomes steeper, and still, inflection points do not exist.

When \(c\) begins at 0 and decreases, it appears to be a slight upward bulge near \(x = 0\). There exist two inflection points for any \(c > 0\).

Step-by-step solution

Identify transitional values of c at which the number of inflection points changes

The family of a curve given by\(f\left( x \right) = {x^4} + cx + x\).

When\(c = 0\)there is no point of inflection, and the curve is a concave upward elsewhere.

When\(c\)increases, then the curve just becomes steeper, and still inflection points do not exist.

When\(c\)begins at 0 and decreases, it appears to be a slight upward bulge near\(x = 0\). There exist two inflection points for any\(c > 0\).

Calculate the second derivative gives the algebraic result as shown below:

\(\begin{array}{c}f'\left( x \right) = 4{x^3} + 2cx + 1\\f''\left( x \right) = 12{x^2} + 2c\end{array}\)

For\(c > 0\),\(f''\left( x \right) > 0\)and for\(c < 0\), there is an inflection point as\(x = \pm \sqrt { - \frac{1}{6}c} \).

When\(c = 0\)the graph contains one critical number, which is somewhere around\(x = - 0.6\)at the absolute minimum value, with\(c\)increases, there is no change in the number of critical numbers.

It is observed that when\(c\)decreases from 0, the graph gradually sprouts another minimum to the right of the origin, anywhere between\(x = 1\)and\(x = 2\). Therefore, the maximum is found near\(x = 0\).

After some observation, we discover that at \(c = - 1.5\), there seem to be two critical numbers: the absolute minimum at \(x = - 1\) and there is no extremum with horizontal tangent at around \(x = 0.5\).

Graph several members of the family

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

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(f\left( x \right) = {x^4} + c{x^2} + x\)in the tab.
2. Set the values as\(c = - \frac{3}{2}, - 3, - 5\). Observe the graph of the function.
3. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the functions as shown below:

It is shown in the graph that with any smaller\(c\), there would be 3 critical numbers as\(c = - 3\), and\(c = - 5\).

We can verify this algebraically by computing\(f'\left( x \right) = 4{x^3} + 2cx + 1\).

When we substitute the value of\(c = - 1.5\)in the equation to obtain as shown below:

\(\begin{array}{c}4{x^3} + 2\left( { - 1.5} \right) + 1 = 4{x^3} - 3x + 1\\ = \left( {x + 1} \right){\left( {2x - 1} \right)^2}\end{array}\)

This has a dual solution at \(x = \frac{1}{2}\), showing that the function contains two critical points: \(x = - 1\) and \(x = \frac{1}{2}\), as we predicted by the graph.