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

Question

1912 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-12T14:04:01

enter image source here

please follow the picture and i pointed 2 in the graph my mistake which should be x=3 .

Answer 2 · 2018-05-12T14:09:58

The function is increasing at #x=3#

Calculate #f'(x)# and lok at the sign of #f'(3)#

#f(x)=e^x(x^2-x)#

The derivative is calculated with the product rule.

#(uv)'=u'v+uv'#

#u=e^x#, #=>#, #u'=e^x#

#v=x^2-x#, #=>#, #v'=2x-1#

Therefore,

#f'(x)=e^x(x^2-x)+e^x(2x-1)#

#=e^x(x^2-x+2x-1)#

#=e^x(x^2+x-1)#

When #x=3#

#f'(3)=e^3(3^2+3-1)=11e^3#

#f'(3)>0#

The function is increasing at #x=3#

graph{e^x(x^2-x) [-6.24, 6.244, -3.12, 3.12]}