Thread: linear to linear fxns

I think I have got a good handle on how to manipulate linear to linear fxns but this is totally stumping me. Tell me what I'm doing wrong?

For each of the following, find the linear function f(x) satisfying the given requirements.
1. f(0)=0
2. f(10)=10
3. f(20)=16

I am taking f(0) to mean y=0 when x=0
f(10)=10 when x=10 and y=10
f(20)=16 when x=20 and y=16

So I enter the points in y=ax+b/x+d and solve for a,b, and d.... right?

Do you need 1 function to meet all 3 requirements (impossible) or do you need 3 functions, each that meets a single requirement.

The wording "find the linear function" if the second case is incorrect. There will be infinitely many linear functions that meet only one these requirements.

For example the first one

$f(0)=0$ is solved by $y=m x$ for any $m \in \mathbb{C}$

So what is the problem really asking?

Sorry, the directions are "find the linear to linear function satisfying the given requirements"

I think the part I'm doing wrong is subtracting the equations to make it a two-variable problem. I am pretty terrible at algebra.

u can use method of langrange interpolating polynomial
f(x)=((x-10)(x-20)/(-200))*0+((x-0)(x-20)/(-100))*10+((x-0)(x-10)/(200))*16
u can also use method of least squares for a quadratic polynomial but then u would have three eqns with three unknowns a,b,c which would be tedious

Originally Posted by UWstudent

Sorry, the directions are "find the linear to linear function satisfying the given requirements"

I think the part I'm doing wrong is subtracting the equations to make it a two-variable problem. I am pretty terrible at algebra.

Ok, I see now. This is new terminology to me.

A linear to linear function is one such that

$f(x)=\dfrac{ax+b}{x+c}$

plugging your numbers in we get

$0=\dfrac b c$

$10=\dfrac {10a+b}{10+c}$

$16=\dfrac{20a+b}{20+c}$

Solving this you find the linear to linear function

Spoiler:

$f(x)=\dfrac{40x}{x+30}$