You can use the fundamental theorem of calculus to find f'(x) without having to integrate.
d/(dx)( integral_x^(x^2) e^(-t^2) dt) = 2 e^(-x^4) x - e^(-x^2)
this right hand side can be simplified as e^(-x^4 - x^2) (2 e^(x^2) x - e^(x^4))
which has zeros at approximately .430 and 1.258 I would use a calculator or wolfram alpha to find the zeros.
Please ask me if you do not understand how to use the fundamental theorem of calculus here
Yanis K.
tutor
the simplification is kind of hard to explain in this interface, it would be much easier to write on a shared whiteboard than to type it out. Basically the -x^4 and -x^2 are non compatible expressions as if they were two different variables because of the negative sign in front of each. if you had an expression of a^-7 - xa^-5 you could make this a little friendlier by factoring out a^(-7-5) or a^-12
this would leave you with a^-12 (a^5-xa^7) and then the expression a^5-xa^7 is what you could set equal to zero without having to input an expression with negative exponents.
IN this case it is possible to simplify the a^-7-5 but with e^(-x^4-x^2) it is not as easy to simplify and would just be harder to look at as e^((1-x^2)/(x^4))
really if you are using a calculator or computer to find the zeros it is not necessary to do anything but type the expression at the right hand side of my first equation in and find the zeros. Let me know if they are the same as what I put down. I could schedule a session as short as 15 minutes online if that would help.
thanks,
Yanis
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05/12/20
Taha T.
thanks a lot. i know how to use the fundamental theorem of calculus and i understood how you got its answer, but i didn't understand how you simpilified the right hand side and how you got the points 0.430 and 1.258 :( can you please clarify?05/11/20