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.
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]: .$\displaystyle -2(72) + b \:=\:-44 \quad\Rightarrow\quad\boxed{b \:=\:100}$
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 $\displaystyle \lim_{x\to a^+}f'(x)= \lim_{x\to a^-}f'(x)$.