# Thread: How do I make this continuous and differentiable?

1. ## How do I make this continuous and differentiable?

I'm lost since "a" and "b" are in the same formula

But I think if I take the derivative of the first one, I might be able to figure out "a" in the second.

Then I would just need a y value to find b.

2. Originally Posted by Zanderist

I'm lost since "a" and "b" are in the same formula

But I think if I take the derivative of the first one, I might be able to figure out "a" in the second.

Then I would just need a y value to find b.
you're on the right track ...

note that the function has to be continuous, and the derivative also has to be continuous.

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

$\displaystyle \lim_{x \to -2^-} f'(x) = \lim_{x \to -2^+} f'(x) = a$

3. Hello, Zanderist!

Find $\displaystyle a$ and $\displaystyle b$ so that the function: .$\displaystyle f(x) \;=\;\bigg\{\begin{array}{ccc}4x^2-6x^2+8 && x < -2 \\ ax + b && x \geq -2 \end{array}$
. . is both continuous and differentiable.

To be continuous, $\displaystyle f(\text{-}2)$ be equal on "both branches".

. . $\displaystyle f(\text{-}2) \:=\:4(\text{-}2)^3 - 6(\text{-}|2)^2 + 8 \:=\:-44$

. . $\displaystyle f(\text{-}2) \:=\:a(\text{-}2) + b \;=\;-2a + b$

. . Hence: .$\displaystyle -2a + b \:=\:-44$ .[1]

To be differentiable, $\displaystyle f'(\text{-}2)$ must be equal on "both branches."

. . $\displaystyle f'(x) \:=\:12x^2-12x \quad\Rightarrow\quad f'(\text{-}2) \:=\:12(\text{-}2)^2 - 12(\text{-}2) \:=\:72$

. . $\displaystyle f'(x) \:=\:a$

. . Hence: .$\displaystyle \boxed{a \:=\:72}$

Substitute into [1]: .$\displaystyle -2(72) + b \:=\:-44 \quad\Rightarrow\quad\boxed{b \:=\:100}$

4. Originally Posted by Soroban
Hello, Zanderist!

To be continuous, $\displaystyle f(\text{-}2)$ be equal on "both branches".

. . $\displaystyle f(\text{-}2) \:=\:4(\text{-}2)^3 - 6(\text{-}|2)^2 + 8 \:=\:-44$

. . $\displaystyle f(\text{-}2) \:=\:a(\text{-}2) + b \;=\;-2a + b$

. . Hence: .$\displaystyle -2a + b \:=\:-44$ .[1]

To be differentiable, $\displaystyle f'(\text{-}2)$ must be equal on "both branches."

. . $\displaystyle f'(x) \:=\:12x^2-12x \quad\Rightarrow\quad f'(\text{-}2) \:=\:12(\text{-}2)^2 - 12(\text{-}2) \:=\:72$

. . $\displaystyle f'(x) \:=\:a$

. . Hence: .$\displaystyle \boxed{a \:=\:72}$

Substitute into [1]: .$\displaystyle -2(72) + b \:=\:-44 \quad\Rightarrow\quad\boxed{b \:=\:100}$

5. By the way, if f is differentiable at, say, x= a, the derivative is not necessarily continuous there, but it does satisfy the "intermediate value property". That's why we can say that we must have $\displaystyle \lim_{x\to a^+}f'(x)= \lim_{x\to a^-}f'(x)$.