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Is #f(x)=e^x/sqrt(x^2-x)# increasing or decreasing at #x=3#?

1 Answer

Jun 13, 2018

Calculate the derivative via the quotient rule:
#f'(x)=(e^xsqrt(x^2-x)-e^x(2x-1)/(2sqrt(x^2-x)))/(x^2-x)#
#=e^x(x^2-2x+1/2)/(x^2-x)^(3/2)#

If this is positive, then #f# is increasing.

#f'(3)=e^3(9-6+1/2)/(9-3)^(3/2)=7/2e^3 6^(3/2)>0#, because each of its component multiplicative parts is #>0#. So #f(x)# is increasing at #x=3#.

For a sanity check, compare the graph of the function:
graph{(e^x)/sqrt(x^2-x) [-9.26, 14.7, 2, 13.99]}

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