Question

let
f(x)=\begin{cases}4x - 6x^{2}&\text{for }x < 0,\\8x^{2}+4x&\text{for }xgeq0.end{cases}
according to the definition of the derivative, to compute (f(0)), we need to compute the left - hand limit (lim_{x
ightarrow0^{-}}square), which is (square), and the right - hand limit (lim_{x
ightarrow0^{+}}square), which is (square). we conclude that (f(0)) is (square).
note: if a limit or derivative is undefined, enter undefined as your answer.

Explanation:

Step1: Recall left - hand derivative formula

The left - hand derivative $f_{-}'(0)=\lim_{x
ightarrow0^{-}}\frac{f(x)-f(0)}{x - 0}$. Since $x<0$, $f(x)=4x - 6x^{2}$ and $f(0)$ (using the right - hand part of the piece - wise function) $f(0)=8\times0^{2}+4\times0 = 0$. So, $\lim_{x
ightarrow0^{-}}\frac{f(x)-f(0)}{x}=\lim_{x
ightarrow0^{-}}\frac{4x - 6x^{2}-0}{x}$.

Step2: Simplify the left - hand limit

$\lim_{x
ightarrow0^{-}}\frac{4x - 6x^{2}}{x}=\lim_{x
ightarrow0^{-}}(4 - 6x)$. Substituting $x = 0$ into $4-6x$, we get $4$.

Step3: Recall right - hand derivative formula

The right - hand derivative $f_{+}'(0)=\lim_{x
ightarrow0^{+}}\frac{f(x)-f(0)}{x - 0}$. Since $x\geq0$, $f(x)=8x^{2}+4x$ and $f(0) = 0$. So, $\lim_{x
ightarrow0^{+}}\frac{f(x)-f(0)}{x}=\lim_{x
ightarrow0^{+}}\frac{8x^{2}+4x-0}{x}$.

Step4: Simplify the right - hand limit

$\lim_{x
ightarrow0^{+}}\frac{8x^{2}+4x}{x}=\lim_{x
ightarrow0^{+}}(8x + 4)$. Substituting $x = 0$ into $8x + 4$, we get $4$.

Step5: Determine the derivative at $x = 0$

Since $f_{-}'(0)=4$ and $f_{+}'(0)=4$, $f'(0)=4$.

Answer:

The left - hand limit is $\lim_{x
ightarrow0^{-}}\frac{4x - 6x^{2}-0}{x}$, which is $4$. The right - hand limit is $\lim_{x
ightarrow0^{+}}\frac{8x^{2}+4x-0}{x}$, which is $4$. $f'(0)$ is $4$.