# Find all $m$ and $b \in \mathbb{R} \text{ st }f(x)$ is differentiable at $x = 1$

Find all $m$ and $b \in \mathbb{R}$ st $f(x)$ is differentiable at $x = 1$

$$f(x) = \begin{cases} mx + b & \text{for x \le 1} \\[6pt] \tan\left(\frac{x\pi}{4}\right) & \text{for 1<x<2} \end{cases}$$

A couple of things caught me off guard with this. First, I am used to seeing one of the variable disappear when I set $x$ equal to the point where the two parts of the function meet; however, here we get:

$m + b = \tan(\frac{\pi}{4}) = 1$ in order for the function to be continuous at $x = 1$.

For it to be differentiable at $x = 1$

$$m = \frac{\pi}{4\cos^2(\frac{\pi}{4})} = \frac{\pi}{2}$$

$$\therefore m = \frac{\pi}{2},\quad b = \frac{2 - \pi}{2}\ ?$$

I wasn't sure if this was right or if I got all the answers.

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Your answer seems correct. –  TMM Apr 30 '12 at 12:23
From the question it just seems like there are more than one answer. Maybe I'm reading into "Find all $m$ and $b$" more than I should. –  stariz77 Apr 30 '12 at 12:25
I guess the "all"-part just implies that you also show that this single solution is the only solution (which you did). If they wrote "Find a (the) solution..." then you could simply give a solution without showing it is unique. –  TMM Apr 30 '12 at 12:31
There is only one solution because there is only one line that "continues" a function from (0,1) to R: the function's tangent in 1. –  Ekuurh Apr 30 '12 at 15:36