Question

1. Find the intervals on which\(f\)is increasing or decreasing.
2. Find the local maximum and minimum values of\(f\).
3. Find the intervals of concavity and the inflection points

23. \(f\left( x \right) = {x^{\bf{4}}} - {\bf{2}}{x^{\bf{2}}} + {\bf{3}}\)

Short answer

1. The function\(f\)changes from increasing to decreasing at \(x = 0\) and from decreasing to increasing at \(x = - 1\) to \(x = 1\).
2. Thelocal maximum values is \(f\left( 0 \right) = 3\) and local minimum values is \(f\left( { \pm 1} \right) = 2\).
3. The function \(f\) is concave upward on \(\left( { - \infty , - \frac{{\sqrt 3 }}{3}} \right)\) and \(\left( {\frac{{\sqrt 3 }}{3},\infty } \right)\) concave downward on \(\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right)\). The inflection points are \(\left( { \pm \frac{{\sqrt 3 }}{3},\frac{{22}}{9}} \right)\).

Step-by-step solution

Write the concavity test

If the function\(f''\left( x \right) > 0\)on an interval, then the graph of the function\(f\)is said to be concave upwardon that interval.

If the function \(f''\left( x \right) < 0\) on an interval, then the graph of the function \(f\) is said to be concave downward on that interval.

(a) Step 2: Find the solutions of the function

Consider the function \(f\left( x \right) = {x^4} - 2{x^2} + 3\).

Differentiate the function w.r.t. \(x\) twice.

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

Since \(f'\left( x \right) = 0\) then find the solutions of the functions.

\(\begin{aligned}{c}4x\left( {x + 1} \right)\left( {x - 1} \right) = 0\\x = 0,1, - 1\end{aligned}\)

Step 3: Construct the table to check whether the function is decreasing or increasing

Now construct the table as shown below:

Interval	\(x + 1\)	\(4x\)	\(x - 1\)	\(f'\left( x \right)\)	\(f\)
\(x < - 1\)	\( - \)	\( - \)	\( - \)	\( - \)	decreasing on \(\left( { - \infty , - 1} \right)\)
\( - 1 < x < 0\)	\( + \)	\( - \)	\( - \)	\( + \)	increasing on \(\left( { - 1,0} \right)\)
\(0 < x < 1\)	\( + \)	\( + \)	\( - \)	\( - \)	decreasing on \(\left( {0,1} \right)\)
\(x > 1\)	\( + \)	\( + \)	\( + \)	\( + \)	increasing on \(\left( {1,\infty } \right)\)

Thus, From the table \(f\) changes from increasing to decreasing at \(x = 0\) and from decreasing to increasing at \(x = - 1\) to \(x = 1\).

(b) Step 4: Find the local maximum and minimum values

Put\(x = 0\)and, \(x = - 1\) and \(x = 1\) into \(f\left( x \right) = {x^4} - 2{x^2} + 3\).

\(\begin{aligned}{c}f\left( 0 \right) = {x^4} - 2{x^2} + 3\\ = 0 - 0 + 3\\ = 3\end{aligned}\)

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

And

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

Thus, the local maximum value is \(f\left( 0 \right) = 3\) and local minimum value is \(f\left( { \pm 1} \right) = 2\).

(c) Step 5: Find the convexity

Differentiate the function \(f'\left( x \right) = 4x\left( {x + 1} \right)\left( {x - 1} \right)\).

\(\begin{aligned}{c}f''\left( x \right) = \frac{d}{{dx}}\left( {4x\left( {x + 1} \right)\left( {x - 1} \right)} \right)\\ = \frac{d}{{dx}}\left( {4x\left( {{x^2} - 1} \right)} \right)\\ = \frac{d}{{dx}}\left( {4{x^3} - 4x} \right)\\ = 12{x^2} - 4\end{aligned}\)

If \(f''\left( x \right) > 0\) then we have,

\(\begin{aligned}{c}12{x^2} - 4 > 0\\12\left( {{x^2} - \frac{1}{3}} \right) > 0\\12\left( {x + \frac{1}{{\sqrt 3 }}} \right)\left( {x - \frac{1}{{\sqrt 3 }}} \right) > 0\\x < - \frac{1}{{\sqrt 3 }}\;{\rm{or}}\,x > \frac{1}{{\sqrt 3 }}\end{aligned}\)

If \(f''\left( x \right) < 0\) then we have,

\(\begin{aligned}{l} - \frac{1}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}\\ - \frac{{\sqrt 3 }}{3} < x < \frac{{\sqrt 3 }}{3}\end{aligned}\).

Thus, \(f\) is concave upward on \(\left( { - \infty , - \frac{{\sqrt 3 }}{3}} \right)\) and \(\left( {\frac{{\sqrt 3 }}{3},\infty } \right)\) concave downward on \(\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right)\).

Step 6: Find the inflection point

Consider the function\(f\left( x \right) = {x^4} - 2{x^2} + 3\).

As the inflection pointis at \(\left( { \pm \frac{{\sqrt 3 }}{3},f\left( { \pm \frac{{\sqrt 3 }}{3}} \right)} \right)\). Find \(f\left( { \pm \frac{{\sqrt 3 }}{3}} \right)\).

\(\begin{aligned}{c}f\left( { \pm \frac{{\sqrt 3 }}{3}} \right) = {x^4} - 2{x^2} + 3\\ = {\left( { \pm \frac{{\sqrt 3 }}{3}} \right)^4} - 2{\left( { \pm \frac{{\sqrt 3 }}{3}} \right)^2} + 3\\ = \frac{1}{9} - 2\left( {\frac{1}{3}} \right) + 3\\ = \frac{1}{9} - \frac{2}{3} + 3\\ = \frac{{1 - 6 + 27}}{9}\\ = \frac{{22}}{9}\end{aligned}\)

Thus, the inflection points are \(\left( { \pm \frac{{\sqrt 3 }}{3},\frac{{22}}{9}} \right)\).