How do you find what x-values are the graph on #f(x)=(2x-3) / (x^2)# concave up and concave down?

1 Answer
Bill K.
Jun 23, 2015

The graph of this function is concave up when #x>9/2# and concave down when #x<9/2# (the function is undefined at #x=0#).

Explanation:

By the Quotient Rule, the first derivative is #f'(x)=\frac{x^2*2-(2x-3)*2x}{x^4}=\frac{2x-4x+6}{x^3}=(6-2x)/(x^3)#.

Using the Quotient Rule again gives the second derivative

#f''(x)=\frac{x^3*(-2)-(6-2x)*3x^2}{x^6}#

#=\frac{-2x-18+6x}{x^4}=(4x-18)/(x^4)#

Clearly #f''(x)>0# when #x>9/2# and #f''(x)<0# when #x<9/2#, making the graph of #f# concave up when #x>9/2# and concave down when #x<9/2# (and #x!=0#).

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