Finding a quadratic function from 2 points and a derivative

[StJimmy]

The text is in slovenian, so i translated it the best i could.
Function f(x) is continuous on the interval (0,5). Calculate a, b and c, if f(4)=11 and f'(3)=1.

 

[JeffM]

First, because the function is continuous on the interval (0, 5)

[MATH]f(2) = \lim_{x \rightarrow 2^-} (2x + 5) = 9.[/MATH]
Do you understand?

Now you can set up three linear equations with a, b, and c as the unknowns.

Can you do that?

 

[StJimmy]

I can. So you get 4a+2b+c=9 for x=2, 16a+4b+c=11 for x=4 and 6a+b=1 for the derivative. I then changed the last equation into b=1-6a and plugged it in the first two. But then i get the equation out both of them (-8a+c=7) and i can't continue.

 

[JeffM]

I can. So you get 4a+2b+c=9 for x=2, 16a+4b+c=11 for x=4 and 6a+b=1 for the derivative. I then changed the last equation into b=1-6a and plugged it in the first two. But then i get the equation out both of them (-8a+c=7) and i can't continue.

Your equations are correct.

[MATH]6a + b = 1 \implies b = 1 - 6a \implies c = 9 - 4a - 2([/MATH]
[MATH]\therefore 4a + 2b + c = 9 \implies 9 = 4a + 2(1 - 6a) + c = c + 2 - 8a \implies c = 8a + 7.[/MATH]
[MATH]\therefore 16a + 4b + c = 11 \implies 11 = 16a + 4(1 - 6a) + 8a + 7 \implies 11 = 11.[/MATH]
What does that mean? It means that a can equal any number.

So we could say that

[MATH]a = 0 \implies c = 7 \ \& \ b = 1 \implies f(x) = x + 7 \ \& \ f'(x) = 1.[/MATH]
Let's check. f (2) = 2 + 7 = 9. Great. f(4) = 4 + 7 = 11. Great. f'(3) = 1. Great.

Or we could say that

[MATH]a = 23 \implies c = 8 * 23 + 7 = 191 \ \& \ b = -\ 137 \implies f(x) = 23x^2 - 137x + 191 \ \& \ f'(x) = 46x - 137.[/MATH]
Let's check.

f((2) = 4 * 23 - 2 * 137 + 191 = 92 + 191 - 274 = 183 - 274 = 9. Perfect.

f(4) = 16 * 23 - 4 * 137 + 191 = 368 - 548 + 191 = 559 - 548 = 11. On the mark.

f'(3) = 3 * 46 - 137 = 138 - 137 = 1. Just as ordered.

Or in general

[MATH]f(x) = ax^2 + (1 - 6a)x + 8a + 7 \implies f'(x) = 2ax + 1 - 6a, \text { and}[/MATH]
[MATH]f(2) = 4a + 2(1 - 6a) + 8a + 7 = 12a - 12a + 2 + 7 = 9, \text { and}[/MATH]
[MATH]f(4) = 16a + 4(1 - 6a) + 8a + 7 = 24a - 24a + 4 + 7 = 11, \text { and }[/MATH]
[MATH]f'(3) = 3 * 2a + 1 - 6a = 6a - 6a + 1 = 1.[/MATH]

 

[StJimmy]

Thank you.