CPM Homework Help : APCALC Problem 4-162

Determine the value of $a$ such that $f$ is continuous at $x = 3$ .

$f ( x ) = \left\{ \begin{array} { c c } { | 4 - 3 x | } & { \text { for } x < 3 } \\ { a x ^ { 2 } + 2 } & { \text { for } x \geq 3 } \end{array} \right.$

Hint:

We need to find a value of $a$ such that:

$\lim \limits_{x\rightarrow 3^{-}} f(x)=\lim \limits_{x\rightarrow 3^{+}} f(x)$

Step 1:

Substitute the left piece of $f\left(x\right)$ and the right piece of $f\left(x\right)$ .

$\lim \limits_{x\rightarrow 3^{-}} \left | 4-3x \right |=\lim \limits_{x\rightarrow 3^{+}} ax^{2}+2$

Step 2:

Evaluate the limits:

$5 = 9a + 2$

Answer:

Solve for $a$ :

$a=\frac{1}{3}$