Question

Find the intervals on which\(f\)is increasing or decreasing, and find the local maximum and minimum values of\(f\).

\(f\left( x \right) = x + \frac{{\bf{4}}}{{{x^2}}}\)

Short answer

The function \(f\) changes from decreasing to decreasing at \(x = 2\). The local minimum value is \(f\left( 2 \right) = 3\).

Step-by-step solution

Write the formula increasing and decreasing test

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

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

Step 2: Find the solutions of the function

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

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

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

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

\(\begin{array}{c}\frac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{{x^3}}} = 0\\x = 0,2\end{array}\)

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

Now construct the table as shown below:

Interval	\({x^3}\)	\(x - 2\)	\(f'\left( x \right)\)	\(f\)
\(x < 0\)	\( - \)	\( - \)	\( + \)	increasing on \(\left( { - \infty ,0} \right)\)
\(0 < x < 2\)	\( + \)	\( - \)	\( - \)	decreasing on \(\left( {0,2} \right)\)
\(x > 2\)	\( + \)	\( + \)	\( + \)	increasing on \(\left( {2,\infty } \right)\)

As \(x = 0\) is not in the domain of the function \(f\) and \(f\) changes from decreasing to decreasing at \(x = 2\).

Step 4: Find local maxima and minima

Put\(x = 2\)into \(f\left( x \right) = x + \frac{4}{{{x^2}}}\).

\(\begin{array}{c}f\left( 2 \right) = 2 + \frac{4}{{{{\left( 2 \right)}^2}}}\\ = 3\end{array}\)

Thus, the local minimum value is \(f\left( 2 \right) = 3\).