What are the extrema of #f(x) = (3x) / (x² - 1)#?

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Answer 1 · 2016-01-02T16:44:35

The function contains no extrema.

Find #f'(x)# through the quotient rule.

#f'(x)=((x^2-1)d/dx(3x)-3xd/dx(x^2-1))/(x^2-1)^2#

#=>(3(x^2-1)-3x(2x))/(x^2-1)^2#

#=>(-3(x^2+1))/(x^2-1)^2#

Find the turning points of the function. These occur when the derivative of the function equals #0#.

#f'(x)=0# when the numerator equals #0#.

#-3(x^2+1)=0#
#x^2+1=0#
#x^2=-1#

#f'(x)# is never equal to #0#.

Thus, the function has no extrema.

graph{(3x)/(x^2-1) [-25.66, 25.66, -12.83, 12.83]}