1. ## Piecewise Differentiability

Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.

2. Originally Posted by Velvet Love
Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.
This will work if the two pieces are equal at $\displaystyle x=2$ (so that $\displaystyle f$ is continuous) and if their derivatives are equal at $\displaystyle x=2$.

Spoiler:
We need to solve these two equations:

1.) $\displaystyle a\cdot 2+b=2-2^2 \implies 2a+b=-2$
2.) $\displaystyle a=-2\cdot 2 \implies a=-4$

Plug $\displaystyle a=-4$ back into the first equation to find that $\displaystyle b=6$.

So $\displaystyle f(x)=\left\{\begin{array}{lr}-4x+6&:x\leq 2\\-x^2+2&:x>2\end{array}\right\}$

3. I actually solved this problem a couple of minutes ago.
Now i'm having trouble with this:
Find the line which is perpendicular to the graph of f(x)=x^2+1 at x=4.

You will need to use the definition to compute the derivative of f.

4. Originally Posted by Velvet Love
Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.
first, remember that if a function is differentiable at $\displaystyle x = c$ , then it is continuous at $\displaystyle x = c$.

secondly ...

$\displaystyle f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$

$\displaystyle f'(2)^- = f'(2)^+$

$\displaystyle \lim_{x \to 2^-} \frac{(ax+b) - f(2)}{x-2} = \lim_{x \to 2^+} \frac{(2-x^2) - f(2)}{x-2}$

5. Nevermind. I found that one out too.
But this one is impossible :/

Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.

You will need to use the definition of the derivative to find the slope of the tangent line.

6. Originally Posted by Velvet Love
Nevermind. I found that one out too.
But this one is impossible :/

Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.

You will need to use the definition of the derivative to find the slope of the tangent line.

$\displaystyle f'(-3) = \lim_{x \to -3} \frac{\frac{1}{x+4} - 1}{x - (-3)}$

start by combining $\displaystyle \frac{1}{x+4} -1$ into a single fraction.

btw ... in future, start a new problem with a new thread.

7. Originally Posted by Velvet Love
Nevermind. I found that one out too.
But this one is impossible :/

Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.

You will need to use the definition of the derivative to find the slope of the tangent line.

$\displaystyle f'(x)=-\frac{1}{(x+4)^2}$. Thus $\displaystyle f'(-3)=-1$. Therefore the slope of the perpendicular line will be $\displaystyle 1$, so you want the equation of the line with slope $\displaystyle 1$ going through the point $\displaystyle (-3,1)$.
$\displaystyle y-1=1\cdot(x-(-3)) \implies \boxed{y=x+4}$
$\displaystyle \frac{1}{x+4} - \frac{x+4}{x+4} = \frac{1-(x+4)}{x+4}$