Question

quadratic inverse
let $f(x)=3(x+1)^2-6$.
17 multiple choice 6 points
for which values of x will $f(x)$ be invertible?
$\bigcirc$ $x \geq -1$
$\bigcirc$ $x \geq -2$
$\bigcirc$ $x \leq 1$
$\bigcirc$ $x \geq -6$
18 multiple choice 6 points
find the inverse.
$\bigcirc$ $3(x-1)^2+6$
$\bigcirc$ $\sqrt{x+2}-1$
$\bigcirc$ $\left(\frac{x+6}{3}\
ight)^2-1$
$\bigcirc$ $\sqrt{\frac{x+6}{3}}-1$

Explanation:

Step1: Analyze invertibility of quadratic

The function $f(x)=3(x+1)^2-6$ is a parabola opening upwards with vertex at $x=-1$. A quadratic is invertible on intervals where it is strictly increasing or decreasing. For this parabola, it is decreasing when $x \leq -1$ and increasing when $x \geq -1$. From the options, $x \geq -1$ is a valid interval for invertibility.

Step2: Swap x and y for inverse

Let $y=3(x+1)^2-6$. Swap $x$ and $y$:
$x=3(y+1)^2-6$

Step3: Isolate the squared term

Add 6 to both sides:
$x+6=3(y+1)^2$
Divide by 3:
$\frac{x+6}{3}=(y+1)^2$

Step4: Solve for y

Take square root of both sides:
$\sqrt{\frac{x+6}{3}}=y+1$
Subtract 1 from both sides:
$y=\sqrt{\frac{x+6}{3}}-1$
(We use the positive root for the interval $x \geq -1$)

Answer:

Question 17:

A. $x \geq -1$

Question 18:

D. $\sqrt{\frac{x+6}{3}} - 1$