Esteban MontielStudios

2022-04-08

Find the constant a such that the function is continuous on the entire real number line.

$f\left(x\right)=\left\{\begin{array}{ll}{x}^{3}& x\le 9\\ a{x}^{2}& x>9\end{array}$

Vasquez

We want the function $f\left(x\right)$ to be continuous on the entire real number line. This means that the left and right limits of $f\left(x\right)$ at $x=9$ must exist and be equal, and the value of $f\left(x\right)$ at $x=9$ must be the same as the limit.
First, let's find the left limit of $f\left(x\right)$ at $x=9$:
${lim}_{x\to {9}^{-}}f\left(x\right)={lim}_{x\to {9}^{-}}{x}^{3}={9}^{3}=729$
Next, let's find the right limit of $f\left(x\right)$ at $x=9$:
${lim}_{x\to {9}^{+}}f\left(x\right)={lim}_{x\to {9}^{+}}a{x}^{2}=a×{9}^{2}=81a$
Since $f\left(x\right)$ is continuous at $x=9$, the left and right limits must be equal:
${lim}_{x\to {9}^{-}}f\left(x\right)={lim}_{x\to {9}^{+}}f\left(x\right)$
$729=81a$
Solving for $a$, we get:
$a=\frac{729}{81}=9$
Therefore, the constant $a$ that makes $f\left(x\right)$ continuous on the entire real number line is $a=9$. So the function becomes:
$f\left(x\right)=\left\{\begin{array}{cc}{x}^{3}\hfill & x\le 9\hfill \\ 9{x}^{2}\hfill & x>9\hfill \end{array}$

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