Question

The figure shows graphs (in blue) of several members of the family of polynomials \(f\left( x \right) = c{x^4} - 4{x^2} + 1\).

1. For which values of c does the curve have minimum points?
2. Show that the minimum and maximum points of every curve in the family lie on the parabola \(y = 2{x^2} + 1\) (shown in red). Identify any transitional values of c where the basic shape change. What happens to the maximum and minimum points and inflection points as c changes? Illustrate by graphing several members of the family.
3. 
   

Short answer

1. When\(c \le 0,\)then\(x = 0\)is the only real solution of\(f'\left( x \right) = 0\). The function\(f'\)varies from positive to negative at\(x = 0\). Therefore,\(f\)contains only a maximum point in this case.

When\(c > 0\), then\(f'\)varies from negative to positive at\(x = \pm \frac{{\sqrt 2 }}{{\sqrt c }}\).

2. Every curve in the family \(f\left( x \right) = c{x^4} - 4{x^2} + 1\) has the minimum points at \(\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}, - \frac{4}{c} + 1} \right)\) , and the maximum point \(\left( {0,1} \right)\) lies on the parabola \(y = - 2{x^2} + 1\).

Step-by-step solution

Determine the value of c does the curve have minimum points

The family of a polynomial given by\(f\left( x \right) = c{x^4} - 4{x^2} + 1\).

Obtain the derivative of the function as shown below:

\(\begin{array}{c}f'\left( x \right) = \frac{d}{{dx}}\left( {c{x^4} - 4{x^2} + 1} \right)\\ = 4c{x^3} - 8x + 0\\ = 4c{x^3} - 8x\\ = 4x\left( {c{x^2} - 2} \right)\end{array}\)

When\(c \le 0,\)then\(x = 0\)is the only real solution of\(f'\left( x \right) = 0\). The function\(f'\)varies from positive to negative at\(x = 0\). Therefore,\(f\)contains only a maximum point in this case.

When\(c > 0\), then;

\(\begin{array}{c}f'\left( x \right) = 4x\left( {c{x^2} - 2} \right)\\ = 4x\left( {\sqrt c x + \sqrt 2 } \right)\left( {\sqrt c x - \sqrt 2 } \right)\end{array}\)

It is observed that \(f'\) varies from negative to positive at \(x = \pm \frac{{\sqrt 2 }}{{\sqrt c }}\).

Show that the maximum and minimum point of every curve and identify the transitional value of c

Substitute\(x = \pm \frac{{\sqrt 2 }}{{\sqrt c }}\)in the function as shown below:

\(\begin{array}{c}f\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}} \right) = c{\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}} \right)^4} - 4{\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}} \right)^2} + 1\\ = \frac{4}{c} - \frac{8}{c} + 1\\ = - \frac{4}{c} + 1\end{array}\)

Substitute\(x = \pm \frac{{\sqrt 2 }}{{\sqrt c }}\)for\(y = g\left( x \right) = - 2{x^2} + 1\)to obtain as shown below:

\(\begin{array}{c}g\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}} \right) = - 2{\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}} \right)^2} + 1\\ = - \frac{4}{c} + 1\end{array}\)

Moreover,\(f\left( 0 \right) = 1\)and\(g\left( 0 \right) = 1\).

Therefore, every curve in the family \(f\left( x \right) = c{x^4} - 4{x^2} + 1\) has the minimum points at \(\left( { \pm \frac{{\sqrt 2 }}{{\sqrt c }}, - \frac{4}{c} + 1} \right)\) , and the maximum point \(\left( {0,1} \right)\) lies on the parabola \(y = - 2{x^2} + 1\).