Piecewise Differentiability

**Velvet Love** · September 17th 2009, 06:08 PM

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.

**redsoxfan325** · September 17th 2009, 06:35 PM

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 $x=2$ (so that $f$ is continuous) and if their derivatives are equal at $x=2$ .

Spoiler:

**Velvet Love** · September 17th 2009, 06:39 PM

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.

**skeeter** · September 17th 2009, 06:40 PM

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 $x = c$ , then it is continuous at $x = c$ .

secondly ...

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

for your piecewise function ...

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

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

**Velvet Love** · September 17th 2009, 07:06 PM

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.

Put your answer in the form y=mx+b.

**skeeter** · September 17th 2009, 07:10 PM

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.

Put your answer in the form y=mx+b.

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

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

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

**redsoxfan325** · September 17th 2009, 07:12 PM

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.

Put your answer in the form y=mx+b.

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

$y-1=1\cdot(x-(-3)) \implies \boxed{y=x+4}$

**Velvet Love** · September 17th 2009, 07:28 PM

Alright. I have ((1-x+4)/(x+4))/(x+3)

**skeeter** · September 18th 2009, 03:37 AM

Originally Posted by Velvet Love

Alright. I have ((1 - x + 4)/(x+4))/(x+3)

you have made a common algebra error ... try again.

$\frac{1}{x+4} - \frac{x+4}{x+4} = \frac{1-(x+4)}{x+4}<br />$

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