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

  • Thread starter KayVee
  • Start date
  • Tags
    Function
In summary, the conversation discusses the difficulties in calculating a function given three coordinates, with one of them being f'(2)=1. The person attempted to use a differential equation, but ultimately found success by using the equation f(x)=ax^2+bx+c and plugging in the given values for f and f'.
  • #1
KayVee
16
0

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:
Physics news on Phys.org
  • #2
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.
 
  • #3
Thanks union68, that worked just fine :smile:
 

1. What is the purpose of finding a function with two coordinates and one f'(x) coordinate?

The purpose of finding a function with two coordinates and one f'(x) coordinate is to determine the equation of the function that passes through the given coordinates and has a derivative of f'(x) at those points. This can help in solving various mathematical problems and understanding the relationship between the function and its derivative.

2. How do you find the function with two coordinates and one f'(x) coordinate?

To find the function with two coordinates and one f'(x) coordinate, you can use the slope-intercept form of a line. First, calculate the slope using the given coordinates and f'(x) coordinate. Then use the slope and one of the coordinates to find the y-intercept. Finally, write the equation in the form of y = mx + b, where m is the slope and b is the y-intercept.

3. Can you find the function with two coordinates and one f'(x) coordinate if the given coordinates are not on the same line?

No, it is not possible to find the function with two coordinates and one f'(x) coordinate if the given coordinates are not on the same line. This is because the derivative of a function represents the slope of the tangent line at a specific point, and if the given coordinates are not on the same line, there is no unique tangent line that can pass through them.

4. Is it necessary for the given coordinates to be distinct in order to find the function with two coordinates and one f'(x) coordinate?

Yes, it is necessary for the given coordinates to be distinct in order to find the function with two coordinates and one f'(x) coordinate. This is because if the coordinates are not distinct, the slope between them will be undefined, and therefore, the function cannot be determined.

5. How can finding a function with two coordinates and one f'(x) coordinate be useful in real-life applications?

Finding a function with two coordinates and one f'(x) coordinate can be useful in various real-life applications, such as physics, engineering, and economics. For example, in physics, the velocity and acceleration of an object can be represented by the derivative and the function, respectively. In economics, the marginal cost and revenue can be expressed as the derivative and function, respectively, which can aid in making business decisions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
722
  • Calculus and Beyond Homework Help
Replies
1
Views
173
  • Calculus and Beyond Homework Help
Replies
8
Views
348
  • Calculus and Beyond Homework Help
Replies
9
Views
916
  • Calculus and Beyond Homework Help
Replies
10
Views
833
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
636
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
404
Back
Top