Question

1. for what value of the constant c is the function ( f(x) ) continuous on ( (-infty, infty) )? f(x) = \begin{cases} cx^2 + 2x & \text{if } x < 2 \\ x^3 - cx & \text{if } x geq 2 end{cases}

Explanation:

Step1: Recall the definition of continuity at a point

For a function \( f(x) \) to be continuous at \( x = a \), the following must hold: \( \lim_{x \to a^{-}} f(x)=\lim_{x \to a^{+}} f(x)=f(a) \). Here, we need to check continuity at \( x = 2 \) (since the function is piece - wise and the only possible point of discontinuity is at \( x = 2 \) where the definition of the function changes).

First, find the left - hand limit as \( x\to2^{-} \) (i.e., \( x \) approaches 2 from the left, so we use the formula \( f(x)=cx^{2}+2x \) for \( x < 2 \)):
\( \lim_{x\to2^{-}}f(x)=\lim_{x\to2}(cx^{2}+2x) \)
Substitute \( x = 2 \) into \( cx^{2}+2x \):
\( \lim_{x\to2^{-}}f(x)=c(2)^{2}+2(2)=4c + 4 \)

Step2: Find the right - hand limit as \( x\to2^{+} \)

For \( x\geq2 \), \( f(x)=x^{3}-cx \). So, the right - hand limit as \( x\to2^{+} \) is:
\( \lim_{x\to2^{+}}f(x)=\lim_{x\to2}(x^{3}-cx) \)
Substitute \( x = 2 \) into \( x^{3}-cx \):
\( \lim_{x\to2^{+}}f(x)=2^{3}-c(2)=8 - 2c \)

Step3: Use the continuity condition

Since \( f(x) \) is continuous at \( x = 2 \), the left - hand limit and the right - hand limit must be equal, i.e.,
\( 4c + 4=8 - 2c \)
Add \( 2c \) to both sides of the equation:
\( 4c+2c + 4=8-2c + 2c \)
\( 6c+4 = 8 \)
Subtract 4 from both sides:
\( 6c+4 - 4=8 - 4 \)
\( 6c=4 \)
Divide both sides by 6:
\( c=\frac{4}{6}=\frac{2}{3} \)

Answer:

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