Finding the function with two coordiantes and one f'(x) cord.

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SUMMARY

The discussion centers on solving for a function given specific coordinates and a derivative value, specifically f'(2)=1, f(0)=3.5, and f(5)=6. The user initially struggled to transform the derivative into a usable function format but found clarity through the suggestion to use the quadratic form f(x)=ax²+bx+c. By substituting the known values into both f(x) and its derivative f'(x), the user was able to establish three linear equations to solve for the unknown coefficients a, b, and c.

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Homework Statement



I'm having troubles calculating a function, because one of the given data is: [tex]f'(2)=1[/tex]

The other two are f(0)=3.5 and f(5)=6.

If three "normal" coordinates are given, i.e (0 ; 3,5) (5 ; 6) and the last one was (2 ; 1), there would be no problem. But how can I transform the f'(2)=1 to a f(x)=y?

Homework Equations



If there were three normal coordinates, I would use the equation: [tex]f(x)=a*x^2+b*x+c[/tex]

The Attempt at a Solution



Since this is the first time I experience the given data, I have no equations for the problem. But I tried to make a differential equation of the function and setting it to zero, but nothing helped.

P.S
I'm still not familiar with the Latex function, since this is my first (or second) homework post. But I need to learn it. Hope you can understand my problem.
 
Last edited:
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You were right with your assumption that the function should be of the form

[tex]f\left(x\right) = ax^2+bx+c[/tex].

What is [tex]f'\left(x\right)[/tex] then? Plug your values into f and f' and you'll have three unknowns and three linear equations.
 
Thanks union68, that worked just fine :smile:
 

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