Questions: -3 ≤ 3x^2+2<1 What is the maximum value f(2x+x=3x^2. 2 for -3 ≤ x<1 ? A) 5 B) 6 C) 27 D) 29 a is a factor of 16 and b is a factor of 27, which is the least value that ab must be a factor of? 1) 1 2) 6 3) 144 4) 432

-3 ≤ 3x^2+2<1
What is the maximum value f(2x+x=3x^2.
2 for -3 ≤ x<1 ?
A) 5
B) 6
C) 27
D) 29
a is a factor of 16 and b is a factor of 27, which is the least value that ab must be a factor of?
1) 1
2) 6
3) 144
4) 432

Transcript text: $-3 \leq 3 x^{2}+2<1$ What is the maximum value $f\left(2 x+x=3 x^{2}\right.$. 2 for $-3 \leq x<1$ ? A) 5 B) 6 C) 27 ) 29 $a$ is a factor of 16 and $b$ is a factor of 27, which is the least value that $a b$ must be a factor of? 1 6 144 432

Solution

Solution Steps

Step 1: Understand the Problem

We need to find the maximum value for $ x $ given the equation $ x = 3x^2 + 2 $ and the constraint $-3 \leq 3x^2 + 2 < 1$.

Step 2: Set Up the Inequality

Given the constraint $-3 \leq 3x^2 + 2 < 1$, we can split it into two parts:

$-3 \leq 3x^2 + 2$
$3x^2 + 2 < 1$

Step 3: Solve the First Inequality

Solve $-3 \leq 3x^2 + 2$: \[ -3 \leq 3x^2 + 2 \] Subtract 2 from both sides: \[ -5 \leq 3x^2 \] Divide by 3: \[ -\frac{5}{3} \leq x^2 \] Since $ x^2 $ is always non-negative, this inequality is always true.

Step 4: Solve the Second Inequality

Solve $3x^2 + 2 < 1$: \[ 3x^2 + 2 < 1 \] Subtract 2 from both sides: \[ 3x^2 < -1 \] Divide by 3: \[ x^2 < -\frac{1}{3} \] Since $ x^2 $ is always non-negative, this inequality has no real solution.