Question

let $g(x)=\

$$\begin{cases}-6x^2 + x & \\text{if } x < 3 \\\\ a & \\text{if } x = 3 \\\\ 2x + 7 & \\text{if } x > 3\\end{cases}$$

$\
\
a. determine the value of $a$ for which $g$ is continuous from the left at $3$.\
b. determine the value of $a$ for which $g$ is continuous from the right at $3$.\
c. is there a value of $a$ for which $g$ is continuous at $3$?\
\
a. the value of $a$ for which $g$ is continuous from the left at $3$ is $\square$.\
(simplify your answer.)

Explanation:

Step 1: Recall the definition of left - continuity

A function \(y = g(x)\) is continuous from the left at \(x = c\) if \(\lim_{x
ightarrow c^{-}}g(x)=g(c)\).

For \(g(x)\) at \(x = 3\), when \(x
ightarrow3^{-}\) (i.e., \(x\) approaches 3 from the left), we use the part of the piece - wise function where \(x\lt3\), which is \(g(x)=- 6x^{2}+x\). And \(g(3)=a\).

Step 2: Calculate the left - hand limit

We need to find \(\lim_{x
ightarrow3^{-}}(-6x^{2}+x)\).
Substitute \(x = 3\) into the function \(-6x^{2}+x\) (since the function \(-6x^{2}+x\) is a polynomial and polynomials are continuous everywhere, so the left - hand limit is equal to the function value at \(x = 3\) for the left - hand part of the piece - wise function).

\(\lim_{x
ightarrow3^{-}}(-6x^{2}+x)=-6\times(3)^{2}+3\)

First, calculate \((3)^{2}=9\). Then, \(-6\times9=-54\). Then, \(-54 + 3=-51\).

Since for left - continuity at \(x = 3\), \(\lim_{x
ightarrow3^{-}}g(x)=g(3)\), we have \(a=-51\).

Answer:

\(-51\)