L3.2

x = -1

x = 0

x = 2

x = 3; x = -2

$$f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \ 2x + 1 & \text{if } x > 1 \end{cases}$$

$$f(x) = \begin{cases} x^2 & \text{if } x < 1 \ 2x + 1 & \text{if } x \geq 1 \end{cases}$$

$$f(x) = \begin{cases} 2x + 1 & \text{if } x \leq 1 \ x^2 & \text{if } x > 1 \end{cases}$$

$$f(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \ x^2 & \text{if } x \geq 1 \end{cases}$$

Yes, it exists

No, it does not exist

It is undefined

It exists but is not continuous

y = 2x + 1

y = 2x - 1

y = -2x + 1

y = x - 1

Yes. The derivative would be equal to zero since there is a vertical tangent.

No. The derivative would not exist at x = 0 since there is a vertical tangent.

No. The derivative does not exist since $$\lim_{x\rightarrow0}\ x^{\frac{1}{3}}$$ does not exist.

Yes since $$\lim_{h\rightarrow0}\ \frac{\left(x+h\right)^{\frac{1}{3}}-x^{\frac{1}{3}}}{h}$$ exists.

Tangent line is vertical

Tangent line has a positive slope

Tangent line is horizontal

Tangent line has a negative slope