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

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Answer 1 · 2016-06-25T06:32:24

#f(x)=e^xsqrt(x^2-x)# is increasing at #x=3#

A function #f(x)# is increasing at #x=a# if #f'(a)>0# and is decreasing at #x=a# if #f'(a)<0#.

As #f(x)=e^xsqrt(x^2-x)#, using product rule

#f'(x)=e^xsqrt(x^2-x)+e^x xx1/(2sqrt(x^2-x))xx(2x-1)#

= #e^xsqrt(x^2-x)+(e^x(2x-1))/(2sqrt(x^2-x))#

and at #x=3#

#f'(x)=e^3sqrt(3^2-3)+(e^3(2xx3-1))/(2sqrt(3^2-3))#

= #e^3sqrt6+(5e^3)/(2sqrt6)=e^3sqrt6(1+5/12)=e^3sqrt6xx17/12#

which is clearly positive.

Hence, #f(x)=e^xsqrt(x^2-x)# is increasing at #x=3#.