How do you find the critical numbers for #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) = (x+3)/x^2# has a single critical point in #x=-6# where it has a local minimum.

Evaluate the derivative of the function:

#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]}