Find a and b such that f(x) is differentiable everywhere

[jwpaine]

Let \(\displaystyle \L f(x) = [\, \begin{eqnarray*} x^2-2x+1, \,\,x <= -2
\\ ax+b, \,\,x > -2
\end{eqnarray*}\)

Find a and b such that \(\displaystyle \L f\) is differentiable everywhere

Do I differentiate each one and then set them equal, and then for a? and then solve for b?

do I sub in -2 into each first? I need some help but not the answer :!:

Thanks!
John

 

[pka]

The function must be continuous at x=-2 so
\(\displaystyle \lim _{x \to - 2^ - } f(x) = \lim _{x \to - 2^ + } f(x)\).

And the righthand and left hand derivatives must be the same.

 

[jwpaine]

The function must be continuous at x=-2 so
\(\displaystyle \lim _{x \to - 2^ - } f(x) = \lim _{x \to - 2^ + } f(x)\).

And the righthand and left hand derivatives must be the same.

Ok so I have:

\(\displaystyle \L \lim _{x \to - 2^ - } f(x)\,\, =\,\,lim _{x \to - 2^ } \,\, x^2-2x + 1 \,\,= [9]\)

\(\displaystyle \L \lim _{x \to - 2^ + } f(x)\,\,=\,\,lim _{x \to - 2^ }\,\, ax + b \,\,= [-2a + b]\)

\(\displaystyle 9 = -2a + b\)

And for the derivatives of each:

2x - 2 = a

Now I have two equations with three variables? I'm missing an important step...... or mainly from lack of understanding :!:

JP

 

[pka]

\(\displaystyle \L f'\left( { - 2^ - } \right) = - 6\quad \& \quad f'\left( { - 2^ + } \right) = a\)

The left derivative must equal the right derivative.

 

[jwpaine]

Oh... of course , because it's being evaluated at -2. Thanks Pka!

so...

\(\displaystyle \L f{'}(-2^{-}) = \lim_{x\to\ -2}\, \frac{f(x) - f(-2)}{x - 2}\)

= \(\displaystyle \L \lim_{x\to\ -2}\, \frac{x^2-2x+1 - 9}{x - 2}\)

= \(\displaystyle \L \lim_{x\to\ -2}\, (x - 4) = (-2-4) = -6\)



But I don't get how \(\displaystyle \L f^{'}(-2^{+}) = a\) :?:

It would seem that the derivative of f(x) = ax + b would = 0

How do I show that \(\displaystyle \L f^{'}(-2^{+}) = a\)?

 

[pka]

How do I show that \(\displaystyle \L f^{'}(-2^{+}) = a\)?

If \(\displaystyle x > -2\) the function is linear.
The derivative of a linear function is the slope of that function.

 

[jwpaine]

Understood. I appreciate you taking the time to help me

John

 

[skeeter]

But I don't get how \(\displaystyle \L f^{'}(-2^{+}) = a\) :?:

It would seem that the derivative of f(x) = ax + b would = 0 ???

if f(x) = ax+b, then f'(x) = a