Question

1. -/6.25 points for what value of the constant c is the function f continuous on (-∞, ∞)? $f(x)=\begin{cases}cx^{2}+2x &\text{if }x < 3\\x^{3}-cx&\text{if }xgeq3end{cases}$ $c = square$

Explanation:

Step1: Recall continuity condition

A function $f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here $a = 3$. Calculate the left - hand limit.
$\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}(cx^{2}+2x)$.
Substitute $x = 3$ into $cx^{2}+2x$: $c(3)^{2}+2\times3=9c + 6$.

Step2: Calculate the right - hand limit

Calculate $\lim_{x
ightarrow3^{+}}f(x)=\lim_{x
ightarrow3^{+}}(x^{3}-cx)$.
Substitute $x = 3$ into $x^{3}-cx$: $3^{3}-c\times3=27 - 3c$.

Step3: Set left - hand and right - hand limits equal

Since the function is continuous at $x = 3$, we set $\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)$.
So, $9c + 6=27 - 3c$.

Step4: Solve for c

Add $3c$ to both sides: $9c+3c + 6=27-3c + 3c$, which gives $12c+6 = 27$.
Subtract 6 from both sides: $12c+6 - 6=27 - 6$, so $12c=21$.
Divide both sides by 12: $c=\frac{21}{12}=\frac{7}{4}$.

Answer:

$\frac{7}{4}$