Is #f(x)=(1-e^x)/x^2# increasing or decreasing at #x=2#?

2 Answers
Oct 28, 2017

The function #f(x)# is decreasing at #x=2#

Explanation:

Calculate the first derivative and look at the sign

We need

#(u/v)'=(u'v-uv')/(v^2)#

Here,

#u(x)=1-e^x#, #=>#, #u'(x)=-e^x#

#v(x)=x^2#, #=>#, #v'(x)=2x#

Therefore,

#f'(x)=(-e^x*x^2-2x(1-e^x))/(x^4)#

#=(-xe^x-2+2e^x)/(x^3)#

When #x=2#

#f'(2)=(-2e^2-2+2e^2)/(8)=-1/4#

As #f'(2)<0#, the function #f(x)# is decreasing at #x=2#

graph{(y-(1-e^x)/(x^2))(y-1000(x-2))=0 [-4.93, 12.85, -4.124, 4.765]}

Oct 28, 2017

#f(x)" is decreasing at x =2"#

Explanation:

#"to determine if a function f(x) is increasing/decreasing at"#
#"x = a evaluate "f'(a)#

#• " if "f'(a)>0" then f(x) is increasing at x = a"#

#• " if "f'(a)<0" then f(x) is decreasing at x = a"#

#"differentiate "f(x)" using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"#

#g(x)=1-e^xrArrg'(x)=-e^x#

#"h(x)=x^2rArrh'(x)=2x#

#rArrf'(x)=(-x^2e^x-2x(1-e^x))/(4x^2)#

#rArrf'(2)=(-4e^2-4+4e^2)/16=-1/4#

#"since "f'(2)<0" then f(x) is decreasing at x = 2"#
graph{(1-e^x)/(x^2) [-10, 10, -5, 5]}