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

1. (Original post by Tallon)
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
Aaaahhh...gosh, yupyup, you're right
2. Spoiler:
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(Original post by Scallym)
y=e^(x^2)

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

dy/dx=0 => x=0
Minima: (0, e^(0^2)) = (0, e^0) = (0,1)
(Original post by Hart92)
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

Cheers guys, realised my mistake

Um, retaking C3 in five days time, not good.
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Retaking 92/100 though, so luckily no serious consequences if I seriously screw it
3. (Original post by ttoby)
If x can take any real number, then x^2 has to be positive or zero, so 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 , giving a final answer of

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 , you should be able to therefore give an answer for c.
Itsn't the range for exponential functions in the form e^x>0 ?
4. (Original post by FZka)
Itsn't the range for exponential functions in the form e^x>0 ?
If x can take any real value then yes, e^x>0. However, x^2 is always positive or zero, and e to the power of a number that's greater than or equal to 0 will be greater than or equal to 1.
5. (Original post by ttoby)
If x can take any real value then yes, e^x>0. However, x^2 is always positive or zero, and e to the power of a number that's greater than or equal to 0 will be greater than or equal to 1.
ohhh. I see.
Missing this point was so stupid of me!

Thanks!

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