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y=e^x+sin(x) How can you show that this equation has real roots? (Maybe with rolle?)
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Roots occur when y=0, the equation transforms to e^x+sin(x)=0
Yes, but how can I proove that there are x's which solve this solution? :)
*this equation
Find y when x=-pi/2 and when x=0 :)
y changes sign between x=-pi/2 and x=0, therefore a root must exist in (-pi/2,0) because this function is continuous
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And this function is continuous because e^x and sin(x) are both continuous
Because if y changes sign in x=[-pi/2,0], y must pass through 0.
that gives me 1 root (with the mean value theorem), but there is an infinite number of roots, right? (because of sin)
Yes, but this is out of the question's scope :)
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