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Stationary Points (e included)

I'm doing C3 Jan 2012 on AQA and currently on Q7a. I got the wrong answer but my method is the same (I have what's on line 2) up until it says that the expression ≠ 0 (line 3, second image)
Screen Shot 2013-04-05 at 10.38.28.png

Screen Shot 2013-04-05 at 10.38.40.png
I don't understand how that helps to figure out the answer?
Original post by PrinceyJ
I'm doing C3 Jan 2012 on AQA and currently on Q7a. I got the wrong answer but my method is the same (I have what's on line 2) up until it says that the expression ≠ 0 (line 3, second image)
Screen Shot 2013-04-05 at 10.38.28.png

Screen Shot 2013-04-05 at 10.38.40.png
I don't understand how that helps to figure out the answer?


Note that the graph of e to the anything is always above the x axis, so it cannot be 0 (so the other bracket must be) :smile:
Reply 2
Avatar for Brister
Brister
7
Original post by PrinceyJ
I'm doing C3 Jan 2012 on AQA and currently on Q7a. I got the wrong answer but my method is the same (I have what's on line 2) up until it says that the expression ≠ 0 (line 3, second image)
Screen Shot 2013-04-05 at 10.38.28.png

Screen Shot 2013-04-05 at 10.38.40.png
I don't understand how that helps to figure out the answer?


a×b=0a \times b = 0 can be satisfied if a=0a = 0 or b=0b = 0 or both.

In this case, eye^y is greater than zero for all real yy so you only get the two solutions from the quadratic. The fact that it does not equal zero does not help to solve the quadratic; it just tells you that you don't need to look for any more solutions.

Note that if you factor out xx from the quadratic, you check that x=0x = 0 works and setting the remaining linear expression equals zero works, too.
Reply 3
Avatar for PrinceyJ
PrinceyJ
OP
12
Original post by Indeterminate
Note that the graph of e to the anything is always above the x axis, so it cannot be 0 (so the other bracket must be) :smile:


Original post by Brister
a×b=0a \times b = 0 can be satisfied if a=0a = 0 or b=0b = 0 or both.

In this case, eye^y is greater than zero for all real yy so you only get the two solutions from the quadratic. The fact that it does not equal zero does not help to solve the quadratic; it just tells you that you don't need to look for any more solutions.

Note that if you factor out xx from the quadratic, you check that x=0x = 0 works and setting the remaining linear expression equals zero works, too.


Thanks guys, I wasn't really in the correct frame of thinking when I came across this

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