2) fx):-5c ーXる (-00,00) f'(x)= 5e-x² .2x = loxe ** O = - 10xe* set =0 x-0(critical point) -plug back into flx) X-0, f(x) = -5 Anything finite divided by 00 = 0 メ=ロ, -lowest %3D check endpoints lim -5e...

How do we know -5 is the absolute min here? Is it because we are decreasing from -infinity until we get to -5 then it goes to +infinity?

2)<br>fx):-5c<br>ーXる<br>(-00,00)<br>f'(x)= 5e-x²<br>.2x<br>= loxe **<br>O = -<br>10xe*<br>set =0<br>x-0(critical point)<br>-plug back into flx)<br>X-0, f(x) = -5<br>Anything finite<br>divided by 00 = 0<br>メ=ロ,<br>-lowest<br>%3D<br>check<br>endpoints<br>lim -5e<br>-5<br>-5<br>verylarge<br>%3D<br>limit, never actually<br>goes to o<br>lim<br>-5e<br>-5<br>-5<br>verylarge<br>Absolute mnof -5 at X=0<br>no absolute max<br>Funcrionn ehnarges concavires<br>

Extracted text: 2) fx):-5c ーXる (-00,00) f'(x)= 5e-x² .2x = loxe ** O = - 10xe* set =0 x-0(critical point) -plug back into flx) X-0, f(x) = -5 Anything finite divided by 00 = 0 メ=ロ, -lowest %3D check endpoints lim -5e -5 -5 verylarge %3D limit, never actually goes to o lim -5e -5 -5 verylarge Absolute mnof -5 at X=0 no absolute max Funcrionn ehnarges concavires

Jun 03, 2022

SOLUTION.PDF

2) fx):-5c ーXる (-00,00) f'(x)= 5e-x² .2x = loxe ** O = - 10xe* set =0 x-0(critical point) -plug back into flx) X-0, f(x) = -5 Anything finite divided by 00 = 0 メ=ロ, -lowest %3D check endpoints lim -5e...

Get Answer To This Question

Related Questions & Answers

Submit New Assignment