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I have equation for tangent line: $y=x+1$ and the tangent line is through point $(x_0,y_0)$ whose coordinate is $(0,1)$ therefore from this formula $m=1=e^x0=e^0$ (the slope of the tangent line) and thus the derivative of the function we are searching for is: $e^x$ to have the function we are searching for,we need to compute $\int e^x~dx= e^x+C$ and we have graph function is through point $(0,1)$ so the constant is just zero therefore the function we are searching for is $e^x$

But is this correct way? How do I know it is the slope of function e^x or some other function?

Thank you so much!

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I'll just give a few things, which you should prove:

Every function staisfying the following two equations would give you a correct solution: $$f(0) = 1 \\ f'(0)=1$$

For example calculate the tangent line at the point $(0,1)$ of the function $f(x) = (x+\frac{1}{2})^2 + \frac{3}{4}$.

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  • $\begingroup$ So in some way: $$tangent-line-from-function\Rightarrow function-from-tangent-line$$ but not the opposite $\endgroup$ – user112637 Dec 1 '13 at 13:20
  • $\begingroup$ Yes that is correct. In fact from the example I think you can get to see the fact that there are infinetly many functions passing trough that point and having that tangent line. $\endgroup$ – Alex Botev Dec 1 '13 at 13:22
  • $\begingroup$ PS: And I would phrase that like this: Given a function you can always get the tangent line everywhere, but given a tangent line you can find inifintely many functions, which are tangent to it a spesific point $\endgroup$ – Alex Botev Dec 1 '13 at 13:23

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