Question

Use the methods of this section to sketch several members of the given family of curves. What do the members have in common? How do they differ from each other?

67. \(f\left( x \right) = {x^4} - cx\), \(c > 0\)

Short answer

As \(c\) increases, the members of the family have one local minimum point with increasing \(x\)-coordinate and decreasing \(y\)-coordinate

Step-by-step solution

Polynomial Functions

The term polynomial function refers to functions that have formed \(f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_{n - 1}}{x^{n - 1}} + {a_n}{x^n}\). Therefore, the polynomial functions of degree \(1,2,3\), and others, depending on the value of \(n\left( {n \in {\mathbb{Z}^ + }} \right)\).

Find the first derivative and second derivative of the function

The given function is \(f\left( x \right) = {x^4} - cx\).

So, first derivation the above function with respect to \(x\) as:

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right) = \frac{d}{{dx}}\left( {{x^4} - cx} \right)\\f'\left( x \right) = \frac{d}{{dx}}\left( {{x^4}} \right) - \frac{d}{{dx}}\left( {cx} \right)\\ = 4{x^3} - c\end{aligned}\)

Now, find the second derivative of the above function as:

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f'\left( x \right)} \right) = \frac{d}{{dx}}\left( {4{x^3} - c} \right)\\f''\left( x \right) = \frac{d}{{dx}}\left( {4{x^3}} \right) - \frac{d}{{dx}}\left( c \right)\\ = 12{x^2}\end{aligned}\)

Increasing or Decreasing Test

Now, by the above first derivative solution, it can be predicted that \(f\left( x \right)\) is increasing when \(f'\left( x \right) > 0\) and decreasing when \(f'\left( x \right) < 0\).

So, \(f\left( x \right)\) is decreasing on \(\left( { - \infty ,{{\left( {\frac{c}{4}} \right)}^{\frac{1}{3}}}} \right)\) and increasing on \(\left( {{{\left( {\frac{c}{4}} \right)}^{\frac{1}{3}}},\infty } \right)\).

Find local maximum

For the continuous function, the local maximum occurs when the graph changes from increasing to decreasing.

The given function is continuous at all points because it is a polynomial. Since the graph is increasing on \(f\left( x \right)\) is decreasing on \(\left( { - \infty ,{{\left( {\frac{c}{4}} \right)}^{\frac{1}{3}}}} \right)\)and increasing on \(\left( {{{\left( {\frac{c}{4}} \right)}^{\frac{1}{3}}},\infty } \right)\).

So, there will be no local maximum because the graph never increases to decrease at any point.

Find local minimum

For the continuous function, the local minimum occurs when the graph changes from decreasing to increasing.

The given function is continuous at all points because it is a polynomial. Since the graph is increasing on \(f\left( x \right)\) is decreasing on \(\left( { - \infty ,{{\left( {\frac{c}{4}} \right)}^{\frac{1}{3}}}} \right)\)and increasing on \(\left( {{{\left( {\frac{c}{4}} \right)}^{\frac{1}{3}}},\infty } \right)\).

So, the local minimum will be at \(x = {\left( {\frac{c}{4}} \right)^{\frac{1}{3}}}\).

Find the interval

As, it can be observed that at\(f''\left( x \right) > 0\)the function \(f\left( x \right)\) is concave up and the function \(f\left( x \right)\) is concave down at \(f''\left( x \right) < 0\).

Therefore, \(f\left( x \right)\) is concave up on \(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

Find the inflection point

Now identify the inflection point of our function. It should be known that inflection occurs when two conditions are met: The graph is continuous at the point, the concavity changes at that point.

So, as it is known that the graph is never concave down, therefore there is no inflection point.

Draw the curve

The member of this family has one local minimum point with increasing \(x\)-coordinate and decreasing \(y\)-coordinate as \(c\) increases.