Question

f(x) = xe^3/x

a) Use l'Hospital's Rule to determine the limit as x 0+.

b)Use calculus to find the local minimum value.

c)Find the interval where the function is concave up.

d) Find the interval where the function is concave down.

Answer 1

e^3???

Answer 2

r u sure thats right?

Answer 3

x*e^(3/x)

Answer 4

first limit is infinity

Answer 5

ok i got that...

Answer 6

assuming it is
\[\lim_{x\rightarrow 0^+}xe^{\frac{3}{x}}\]

Answer 7

derivative via product and chain rule gives
\[\frac{e^{\frac{3}{x}} (x-3)}{x}\] yes?

Answer 8

critical point at x = 3 (also 0 but not in the domain so can ignore that one

Answer 9

so local min at x = 3. point is
\[(3,3e)\]

Answer 10

3 doing work

Answer 11

neither does 3e

Answer 12

i will guess concave up on
\[(0,\infty)\][

Answer 13

?

Answer 14

critical point is x = 3, yes? not sure what you meant. did you mean "3 doesn't work?"

Answer 15

i plugged it into my online homework...its wrong....I got the concave up and down already

Answer 16

what does the homework say exactly? does it want both values? i mean both input and output? because i am fairly certain this is right

Answer 17

just the local minimum

Answer 18

well if that is what is says, the local minimum is
\[3e\]

Answer 19

\[\color{#ff0099}{\text{hello my myininaya!}}\]

Answer 20

did i screw up somewhere? looks good to me

Answer 21

\[f(x)=xe^{\frac{3}{x}}=>f'(x)=1*e^\frac{3}{x}+x*\frac{-3}{x^2}*e^\frac{3}{x}\]
\[e^\frac{3}{x}(1+\frac{-3}{x})=0\]
\[e^\frac{3}{x}\neq0, \frac{x-3}{x}=0=>x=3=>f(3)=3*e^\frac{3}{3}=3*e^1=3e\]
you are right

Answer 22

its 8.15

Answer 23

i had to find the value for e

Answer 24

3e=8.154845 approximately
your question doesn't say round to the hundredths

Answer 25

yes it does on the online problem thanks

Answer 26

ok but you got it wrong
thats weird

Answer 27

or are you didn't round?

Answer 28

i'm confused whats wrong then?

Answer 29

i put 3e in