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Finding the range of a graph

f(x) = ex2
find the range of f(x)
How do I do this - I don't know how to draw this graph?


g(x) = x2 + 3ex2
Again, how is this done without drawing the graph?
(edited 12 years ago)
Reply 1
Avatar for ilovemath
ilovemath
OP
2
so for the first one, the dy/dx = 2xex2 which gives a min. point of 0 BUT THE ANSWER is f(x) >= 1
why?
Reply 2
Avatar for MrGum
MrGum
Original post by ilovemath
f(x) = ex2
find the range of f(x)
How do I do this - I don't know how to draw this graph?


Decide what values x2 can take (using a graph of y=x2 if necessary)
Use a graph of ex to decide what possible outputs are possible when the input is in the set above.


g(x) = x2 + 3ex2
Again, how is this done without drawing the graph?


As above, except now you need to know a bit about the behaviour of x+3ex to get an idea of its graph (you don't need a detailed sketch). Knowledge of the sign of dy/dx for this function will help.
Reply 3
Avatar for MrGum
MrGum
Original post by ilovemath
so for the first one, the dy/dx = 2xex2 which gives a min. point of 0 BUT THE ANSWER is f(x) >= 1
why?


By "a min point of 0" I think you mean a minimum when x=0 - what is the y-co-ordinate of this minimum point? That is really all you need to do an adequate sketch graph. A similar method will work for the other example.
Reply 4
Avatar for ilovemath
ilovemath
OP
2
Original post by Mr Gum
By "a min point of 0" I think you mean a minimum when x=0 - what is the y-co-ordinate of this minimum point? That is really all you need to do an adequate sketch graph. A similar method will work for the other example.


ahh got the first one
for the second one
dy/dx = 2x + 6xex2
which, on factorising shows min occurs when x = 0
so
(0)^2 + 3e^(0^2)) = 3
so g(x) >=3

ok thanks
Reply 5
Avatar for MrGum
MrGum
Original post by ilovemath
ahh got the first one
for the second one
dy/dx = 2x + 6xex2
which, on factorising shows min occurs when x = 0
so
(0)^2 + 3e^(0^2)) = 3
so g(x) >=3

ok thanks


Correct! (did you check the other factor of dy/dx doesn't lead to a second stationary point?)
Reply 6
Avatar for ilovemath
ilovemath
OP
2
Original post by Mr Gum
Correct! (did you check the other factor of dy/dx doesn't lead to a second stationary point?)


yes, it couldn't because ln of a minus number DOES NOT EXIST
Reply 7
Avatar for MrGum
MrGum
Original post by ilovemath
yes, it couldn't because ln of a minus number DOES NOT EXIST


Yes.

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