\(f(x) = 4 - \frac{2}{x}+\frac{6}{x^{2}}\)

Take derivative and find out critical points to get maxima and minima

Apply power rule

\(f(x) = 4 - \frac{2}{x}+\frac{6}{x^{2}}\)

\(f`(x) = 0 + \frac{2}{x^{2}}+\frac{12}{x^{3}}\)

\(f`(x) = \frac{2}{x^{2}}+\frac{12}{x^{3}}\)

Set the derivative \(=0\) and solve for x

LCD is \(x^{3}\),

multiply each term by \(x^{3}\)

\(\frac{2}{x^{2}}-\frac{12}{x^{3}}=0\)

\(\frac{2}{x^{2}}(x^{3})-\frac{12}{x^{3}}(x^{3})=0(x^{3})\)

\(2x - 12 = 0\)

\(2x = 12\)

\(x = 6\)

\(x=0\) makes denominator 0 and it is undefined

So \(x=0, x=6\)

Break the numbers line into three intervals using 0 and 6

Check each interval using derivative Let \(x=-1 , x=1, x=7\)

\(f'(x)=\frac{2}{x^{2}}-\frac{12}{x^{3}}\)

\(f'(-1)=\frac{2}{(1)^{2}}-\frac{12}{(1)^{3}}=-10\)

\(f'(1)=\frac{2}{(1)^{2}}-\frac{12}{(1)^{3}}=-10\)

\(f'(7)=\frac{2}{(7)^{2}}-\frac{12}{(7)^{3}}=\frac{2}{343}\)

Derivative is from positive in two intervals

So increasing intervals are\((-\infty,0) U(6, \infty)\)

Decreasing interval \((0, 6)\)

There is a break in graph of \(f(x) at x = 0\)

The derivative goes from negative to positive at \(x = 6\)

So relative minima at \(x = 6\)

There is no relative maxima

Take derivative and find out critical points to get maxima and minima

Apply power rule

\(f(x) = 4 - \frac{2}{x}+\frac{6}{x^{2}}\)

\(f`(x) = 0 + \frac{2}{x^{2}}+\frac{12}{x^{3}}\)

\(f`(x) = \frac{2}{x^{2}}+\frac{12}{x^{3}}\)

Set the derivative \(=0\) and solve for x

LCD is \(x^{3}\),

multiply each term by \(x^{3}\)

\(\frac{2}{x^{2}}-\frac{12}{x^{3}}=0\)

\(\frac{2}{x^{2}}(x^{3})-\frac{12}{x^{3}}(x^{3})=0(x^{3})\)

\(2x - 12 = 0\)

\(2x = 12\)

\(x = 6\)

\(x=0\) makes denominator 0 and it is undefined

So \(x=0, x=6\)

Break the numbers line into three intervals using 0 and 6

Check each interval using derivative Let \(x=-1 , x=1, x=7\)

\(f'(x)=\frac{2}{x^{2}}-\frac{12}{x^{3}}\)

\(f'(-1)=\frac{2}{(1)^{2}}-\frac{12}{(1)^{3}}=-10\)

\(f'(1)=\frac{2}{(1)^{2}}-\frac{12}{(1)^{3}}=-10\)

\(f'(7)=\frac{2}{(7)^{2}}-\frac{12}{(7)^{3}}=\frac{2}{343}\)

Derivative is from positive in two intervals

So increasing intervals are\((-\infty,0) U(6, \infty)\)

Decreasing interval \((0, 6)\)

There is a break in graph of \(f(x) at x = 0\)

The derivative goes from negative to positive at \(x = 6\)

So relative minima at \(x = 6\)

There is no relative maxima