Applied Calculus MA25C01 | ✅ Continuity Problem4 : \[ f(x) = \begin{cases} cx^2 + 2x, & x < 2 \\ x^3 - cx, & x \geq 2 \end{cases} \]

▶ Tutorial Video:Continuity -Example 4

Example: For what values of constant \( c \) is the function \( f \) continuous on \( (-\infty, \infty) \)?

\[ f(x) = \begin{cases} cx^2 + 2x, & x < 2 \\ x^3 - cx, & x \geq 2 \end{cases} \]

[Ref: N/D-22-R-21-ARR]

Solution

For \( f(x) \) to be continuous at \( x = a \), the following condition must be met:

\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \]

Since both pieces of the function are polynomials, we only need to check continuity at the transition point \( x = 2 \).

1. Left-hand Limit (as \( x \to 2^- \)):

\[ \lim_{x \to 2^-} (cx^2 + 2x) = c(2)^2 + 2(2) = 4c + 4 \]

2. Right-hand Limit and \( f(2) \) (as \( x \to 2^+ \)):

\[ \lim_{x \to 2^+} (x^3 - cx) = (2)^3 - c(2) = 8 - 2c \]

3. Equating the Limits:

For the function to be continuous, the limits must be equal:

\[ 4c + 4 = 8 - 2c \] \[ 4c + 2c = 8 - 4 \] \[ 6c = 4 \] \[ c = \frac{4}{6} \]

Result: \( c = \frac{2}{3} \)

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