# Thread: solving for a continuous function

1. ## solving for a continuous function

let m and b be real numbers and let the function f be defined by
$\displaystyle f(x)=\begin{cases} & \text{ if } x= 1+3bx+2x^2 (x\leq 1) \\ & \text{ if } x= mx+b (x greater than 1) \end{cases}$
If f is both continuous and differentiable at x=1, then
m=?
b=?

my steps
-substitute for x=1
$\displaystyle 1+3b(1)+2(1)^2=m(1)+b$
-combine like terms
$\displaystyle 3+3b=m+b$
-rearrange for m
$\displaystyle m=3+2b$
-substitute back in
3+3b=(3+2b)+b
3+3b=3+3b

-this is where i am lost i dont even have a variable when I solve...
what did I do wrong?

All help is appreciated.

2. Originally Posted by dandaman
let m and b be real numbers and let the function f be defined by
$\displaystyle f(x)=\begin{cases} & \text{ if } x= 1+3bx+2x^2 (x\leq 1) \\ & \text{ if } x= mx+b (x greater than 1) \end{cases}$
If f is both continuous and differentiable at x=1, then
m=?
b=?
If $\displaystyle f$ is continous it means $\displaystyle \lim_{x\to 1^{-}} f(x) = \lim_{x\to 1^{+}} f(x)$.
Thus, $\displaystyle \lim_{x\to 1^{-}} 1 + 3bx+2x^2 = \lim_{x\to 1^+} mx+b$. [1]
If $\displaystyle f$ is differenciable it means $\displaystyle f'(1-) = f'(1+)$ (derivative from left equals derivative from right).
Thus, $\displaystyle \frac{d}{dx}(1+3bx+2x^2)\bigg|_{x=1} = \frac{d}{dx}(mx+b)\bigg|_{x=1}$. [2]

The equations [1] and [2] allow you to solve a system of equations.