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Exact coordinates of stationary point y=e^sinx

I have found the derivative of the function above.

I’m just not sure how to deal with the e part.

Usually I would multiple though by In to cancel out e but it doesn’t work in this case.

See attached

Question 10
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(edited 1 year ago)
Can you solve
e^(sin(x)) = 0
If not, what about the other term that multiplies it?
Reply 2
Original post by mqb2766
Can you solve
e^(sin(x)) = 0
If not, what about the other term that multiplies it?


Cos x=0 would be π/2 and 3π/2.

I’m not sure how to deal with the other part. I suppose I could cancel it out by dividing through by e^sinx leaving cos x by it self hence two x above


Edit: for arguments sake though how would I find the value of e? As I would cancel e and have sin x =0 hence π and or am I wrong in that approach
(edited 1 year ago)
Original post by KingRich
Cos x=0 would be π/2 and 3π/2.

I’m not sure how to deal with the other part. I suppose I could cancel it out by dividing through by e^sinx leaving cos x by it self hence two x above


Sort of for the division. Really you have
f(x)*g(x) = 0
so either (or both) f(x)=0 or g(x)=0 just like your usual linear factors/roots. You get this product when you apply the chain rule.

You can divide through by e^(sin(x)) if its not zero, is it? However, its better to think of factors, rather than division as its easy to forget to check whether its not zero.

Another way to think about/visualise the original function is to think about where sin(x) has stationary points and what effect the e^* has. Sketch it or plot in desmos if necessary. The same sketch should convince you whether its ever zero.
(edited 1 year ago)
Reply 4
Original post by KingRich
I have found the derivative of the function above.

I’m just not sure how to deal with the e part.

Usually I would multiple though by In to cancel out e but it doesn’t work in this case.



It might just be sloppy use of language but you should not be thinking of "multiplying by ln to cancel out e" - it doesn't work like that! You can take logs to cancel out exponentiation because ln x is the inverse of e^x, but in this case you shouldn't need to be thinking along those lines.

What do you know about exponential functions in general? Can axa^x ever be 0 for any base a? Don't forget that ax×ax=a0=1a^x \times a^{-x} = a^0 = 1, so what would axa^{-x} have to be if axa^x were 0?
Original post by KingRich
Cos x=0 would be π/2 and 3π/2.

I’m not sure how to deal with the other part. I suppose I could cancel it out by dividing through by e^sinx leaving cos x by it self hence two x above


Edit: for arguments sake though how would I find the value of e? As I would cancel e and have sin x =0 hence π and or am I wrong in that approach


For your edit, note that e is the usual 2.71828..... but its value is not that important for the question (though its properties are). Its just the usual exponential funtion so
e^sin(x) = exp(sin(x))
(edited 1 year ago)
Reply 6
Original post by mqb2766
Sort of for the division. Really you have
f(x)*g(x) = 0
so either (or both) f(x)=0 or g(x)=0 just like your usual linear factors/roots. You get this product when you apply the chain rule.

You can divide through by e^(sin(x)) if its not zero, is it? However, its better to think of factors, rather than division as its easy to forget to check whether its not zero.

Another way to think about/visualise the original function is to think about where sin(x) has stationary points and what effect the e^* has. Sketch it or plot in desmos if necessary. The same sketch should convince you whether its ever zero.


Mmm, the chain rule is what I applied to find the derivative here.

I need to consider them as two separate x’s then as you say and I’ve made that error before where I miss possible solutions, so what I’m seeing is cosx=0 and e^sinx=0

but cos x=0 has solutions two solutions . Where as e^sin x=0 could have some x solutions too where sin x=0 we have π and this would suggest stationary points at (π,1) and (2π,1) without using desmos, it’s to see e^x effects the graph. Only from there can I see they’re not solutions
(edited 1 year ago)
Original post by KingRich
Mmm, the chain rule is what I applied to find the derivative here.

I need to consider them as two separate x’s then as you say and I’ve made that error before where I miss possible solutions, so what I’m seeing is cosx=0 and e^sinx=0

but cos x=0 has solutions two solutions . Where as e^sin x=0 could have some x solutions too where sin x=0 we have π and this would suggest stationary points at (π,1) and (2π,1) without using desmos, it’s to see e^x effects the graph. Only from there can I see they’re not solutions

exp(x) > 0 for all x,, it can never be zero.

And as sin(x)>=-1 and sin(x)<=1, you have
exp(-1) <= exp(sin(x)) <= exp(1)
where exp(-1) = 1/e ~ 1/3 and exp(1) = e ~3.

So e^sin(x) can never be zero. The only stationary points are when cos(x)=0.

Using the fact that e^* is always positive is an important property.
Reply 8
Original post by mqb2766
exp(x) > 0 for all x,, it can never be zero.

And as sin(x)>=-1 and sin(x)<=1, you have
exp(-1) <= exp(sin(x)) <= exp(1)
where exp(-1) = 1/e ~ 1/3 and exp(1) = e ~3.

So e^sin(x) can never be zero. The only stationary points are when cos(x)=0.

Using the fact that e^* is always positive is an important property.


Right!!! As e^x has c-intercept 1 is proof alone that it cannot equate to 0! Right. :smile:
Original post by KingRich
Right!!! As e^x has c-intercept 1 is proof alone that it cannot equate to 0! Right. :smile:

e^0 = 1
but noting its always positive is a bit more than that.
Tbh its fairly clear from its definition as
e^x
which means a positive number (e=2.71828...) raised to a power, which must be > 0. As per the previous post, its an important property which will come up in one form or another on your exams. Make sure you have the curve
e^x
clear in your head (use desmos if necessary).
(edited 1 year ago)
Reply 10
Original post by mqb2766
e^0 = 1
but noting its always positive is a bit more than that.
Tbh its fairly clear from its definition as
e^x
which means a positive number (e=2.71828...) raised to a power, which must be > 0. As per the previous post, its an important property which will come up in one form or another on your exams. Make sure you have the curve
e^x
clear in your head (use desmos if necessary).


Mmm, I have both e^x and in x like being mirror image through line y=x. Admittedly when I first did these in year 1 I wasn’t 100% clear of what it all meant but it becomes easier to understand the more I have to use the graphs etc
Original post by KingRich
Mmm, I have both e^x and in x like being mirror image through line y=x. Admittedly when I first did these in year 1 I wasn’t 100% clear of what it all meant but it becomes easier to understand the more I have to use the graphs etc


e^x = exp(x) and ln(x) are indeed inverse functions so you can interpret that as being reflections in the line y=x. You know ln(x) is only defined for x>0. This is equivalent (by reflection) to saying e^x > 0. So the domain of ln() is equal to the range of exp() which is the postive numbers in both cases.
I guess youre now sorted, but a slightly different take on the above so trying to solve
e^sin(x) = 0
Taking logs (as davros mentioned, this is not multiplication) which is the inverse of exp() so
ln(e^sin(x)) = sin(x)
so you'd be looking for solutions of
sin(x) = ln(0)
However the right hand side is not defined so there are no solutions. The quick way is to note that e^* > 0, but if you do go through the algebra (correctly) you arrive at the same conclusion.
Reply 13
Original post by mqb2766
I guess youre now sorted, but a slightly different take on the above so trying to solve
e^sin(x) = 0
Taking logs (as davros mentioned, this is not multiplication) which is the inverse of exp() so
ln(e^sin(x)) = sin(x)
so you'd be looking for solutions of
sin(x) = ln(0)
However the right hand side is not defined so there are no solutions. The quick way is to note that e^* > 0, but if you do go through the algebra (correctly) you arrive at the same conclusion.


Yes, thank you

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