Question

1. (13 points) find a number c such that

$f(x) = \

$$\begin{cases} cx^2 + 2x, & \\text{if } x < 2 \\\\ x^3 - cx, & \\text{if } x \\geq 2 \\end{cases}$$

$
is continuous at $x = 2$.

Explanation:

Step1: Compute left limit at x=2

Left limit: $f(2^-) = c(2)^2 + 2(2) = 4c + 4$

Step2: Compute right limit at x=2

Right limit: $f(2^+) = (2)^3 - c(2) = 8 - 2c$

Step3: Set limits equal for continuity

$4c + 4 = 8 - 2c$
Solve for c: $6c = 4$ → $c = \frac{4}{6} = \frac{2}{3}$

Answer:

$\frac{2}{3}$