Who can explain this to me in the simplest way possible? (C3)

How do I do question a) and c)? This is my sole weakness, and it's so simple. I found that subbing x=0 works but my math teacher said it was wrong and then he explained some random stuff that I got lost in about a minute after he started speaking.

So in the simplest possible way, how do I go about doing this question (a and c only)?

thanks

just draw the graph

the range i belive is just what the the y values will be for that function
just look at the graphs and see if it helps. you want to find the values of y for which the function exists

Reply 3

Mr.Singh

this isn't mei is it. ref-4753...?

Reply 4

jimber

9

the range of g(x) is just >0 ?

Reply 5

Dreizhen

13

Yeah, can't do much besides draw graphs for g(x) and fg(x)

Reply 6

Skadoosh

12

Find smallest value of x^2.

Reply 7

ttoby

15

If x can take any real number, then x^2 has to be positive or zero, so

e^{x^2}

will always be e to the power of a number greater than or equal to zero. From what you know about the graph of e^x, you can say that e to the power of any number greater than or equal to zero will be greater than or equal to 1, so

e^{x^2} \geq 1

, giving a final answer of

g \geq 1

Note that since this is a 1 mark question, you can go through this thought process in your head or in rough, and just write down the final answer. You won't be asked for an explanation for your range.

Using what you know about x^2 and

e^{x^2}

, you should be able to therefore give an answer for c.

Reply 8

victor_b1992

10

jimber

the range of g(x) is just >0 ?

no g(x)>or equal to 1

Reply 9

Tallon

16

you know that x^2 is always positive. So what's the smallest value e^u can take when u is positive? Obviously 0, right? so the range is what?

Reply 10

Scallym

Try to compare it to something you're familiar with, and what difference the difference between them makes. (If that made any sense. :p:

)

If f(x)=e^x, f(x)>0 (When x is real)

x^2 is always positive, what difference does does this make to the range of values g(x) can take?

Part c) obviously depends on a), but it's not that hard to see when you have a). (Consider x=0 and as x tends to infinity.)

:smile:

(Sorry for using f(x) when it's already defined in the question, but k(x) or whatever looks odd :tongue:

)

Not quite jiimber, btw.

Reply 11

Scallym

Tallon

you know that x^2 is always positive. So what's the smallest value e^u can take when u is positive? Obviously 0, right? so the range is what?

Not quite.

Reply 12

Tallon

16

Scallym

Not quite.

edit -
meant smallest u can take for e^u.
e^u for u =x^2 = 1 for smallest u

Reply 13

Scallym

Tallon

????

The smallest value is 1.
The limit is 0 as x--> -infinity, but if x is positive, it's 1.

Edit:

Aaah, got you.
Sorry. :tongue:

Reply 14

Hart92

For b) you just sub the equation for g into the equation for f, then you get 3e^x^2 + lne^x^2 Then the ln and e cancel giving you the equation x^2 + 3e^x^2, simples :smile:

Reply 15

goldsilvy

15

Tallon

edit -
meant smallest u can take for e^u.
e^u for u =x^2 = 1 for smallest u

(0.5)^2=0.25

smallest value is definitely not 1. I reckon the range to be g(x)>0

The shape should be as of a simple e^x, i.e. on the negative side, as x tend to minus infinity, the g(x) tend to zero.

Reply 16

Tallon

16

goldsilvy

(0.5)^2=0.25

smallest value is definitely not 1. I reckon the range to be g(x)>0

The shape should be as of a simple e^x, i.e. on the negative side, as x tend to minus infinity, the g(x) tend to zero.

err no.
e^(x^2)
means e is always to the power of a positive.
when the positive is big so is e^(x^2).
the smallest it could be is a number very close to 0
e^0 = 1
so range is g(x) > 1

Reply 17

goldsilvy

15

For part c)

f(x) is conditioned to have x>0, so substitute into equation fg(x) x=0; x=1; x =2. See what you can notice and try to conclude the range.

Reply 18

Hart92

goldsilvy

(0.5)^2=0.25

smallest value is definitely not 1. I reckon the range to be g(x)>0

The shape should be as of a simple e^x, i.e. on the negative side, as x tend to minus infinity, the g(x) tend to zero.

No x^2 is ALWAYS positive so on the graph there are no -x values and as the graph of e passes through the y-axis at 1, 1 is the lowest y value e^x^2 can take therefor g(x)>=1

Reply 19

Scallym

goldsilvy

(0.5)^2=0.25

smallest value is definitely not 1. I reckon the range to be g(x)>0

The shape should be as of a simple e^x, i.e. on the negative side, as x tend to minus infinity, the g(x) tend to zero.

y=e^(x^2)

dy/dx=2xe^(x^2)

dy/dx=0 => x=0
Minima: (0, e^(0^2)) = (0, e^0) = (0,1)

Who can explain this to me in the simplest way possible? (C3)

Related discussions

Latest

Trending

Trending

Articles for you