How do you find the critical numbers for $f (x) = \frac{x + 3}{x^{2}}$ $f(x)= (x+3) / x^2$ to determine the maximum and minimum?

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Answer 1 · 2017-07-19T07:35:20

$f (x) = \frac{x + 3}{x^{2}}$ $f(x) = (x+3)/x^2$ has a single critical point in $x = - 6$ $x=-6$ where it has a local minimum.

Evaluate the derivative of the function:

$f (x) = \frac{x + 3}{x^{2}} = \frac{1}{x} + \frac{3}{x^{2}}$ $f(x) = (x+3)/x^2 = 1/x +3/x^2$

$f'(x) = -1/x^2-6/x^3 = -(x+6)/x^3$

The critical points are the solutions of the equation:

$f'(x) = 0$

$-(x+6)/x^3 = 0$

so the only critical point is for $x= -6$ .

Evaluate the second derivative:

$f''(x) = 2/x^3+18/x^4 = (2x+18)/x^4$

As $f''(-6) > 0$ the point is a local minimum.

graph{(x+3)/x^2 [-10, 2, -0.2, 0]}

How do you find the critical numbers for f(x)= (x+3) / x^2f(x)=x+3x2f(x)= (x+3) / x^2 to determine the maximum and minimum?