There are continuous functions, for example f(t) = e^{t^2}, for which the integral defining the Laplace transform does not converge for any value of the Laplace variable s. So you could say that this continuous function does not have a Laplace transform.
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That's true. If a function is continuous, it's (Riemman) integrable, but the converse is not true.
consumption is that money who you consume on any thing and the consumption function is that relation who tell you the consuming level on your every money income level.
Υou show that it is continuous in every element of it's domain.
No, this is not a function. The graph would have a vertical line at x=-14. Since there are more than one y value for every given x value, the equation does not represent a function. The slope of the equation also does not exist.
No.The equation x/(x^2 + 1) does not have a vertical asymptote.