Differentiability: f(x) = cx+d for x<=2, x^2-cx for x>2

[tjkubo]

If
\(\displaystyle f(x)= \left\{ \begin{array}{c}cx+d, \mbox{ } x\leq 2\\ x^2-cx, \mbox{ } x>2\)
and f'(x) is defined at x=2, what is the value of c+d?

 

[skeeter]

Re: Differentiability

a function is continuous at x = a if ...

1. f(a) is defined.
2. lim{x->a} f(x) exists.
3. lim{x->a} f(x) = f(a)

use this definition of continuity for the functions f(x) and f'(x) ... you'll find c and d.

 

[tjkubo]

Re: Differentiability

I tried that but then I got the equation
\(\displaystyle 2c+d=4-2c\)
which is where I got stuck.

 

[o_O]

Re: Differentiability

The derivative should also be equal at x = 2:
\(\displaystyle f'(x) = \left\{ \begin{array}{ll} c & x \leq 2 \\ 2x - c & x > 2\)

Since f'(x) is defined, the deriative at x = 2 can't simultaneously hold 2 values so you can solve for c and then plug it back into the original equation to solve for d.

 

[skeeter]

Re: Differentiability

I tried that but then I got the equation
\(\displaystyle 2c+d=4-2c\)
which is where I got stuck.

you did it for f(x) ... did you do it for f'(x)?