Question

find the inverse of the following function. also list the domain restriction for the inverse
$f(x) = (x + 3)^2 - 5$, for $x < -3$
$f^{-1}(x) = \boxed{\quad}$ for $\boxed{\quad}$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x) = (x + 3)^2 - 5 \) for \( x < -3 \). Replace \( f(x) \) with \( y \), so we have \( y=(x + 3)^2 - 5 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap \( x \) and \( y \). This gives us \( x=(y + 3)^2 - 5 \).

Step3: Solve for \( y \)

First, add 5 to both sides of the equation: \( x + 5=(y + 3)^2 \).
Then, take the square root of both sides. Since the original function has \( x < -3 \), the original function is decreasing (because the square function \( (x + 3)^2 \) is decreasing for \( x < -3 \)), so the inverse function should have a square root with a negative sign (to get the decreasing part inverse). So we have \( \sqrt{x + 5}=-(y + 3) \) (we take the negative square root because the original function's domain is \( x < -3 \), so the range of the original function will be the domain of the inverse, and we need to get the correct inverse).
Now, solve for \( y \):
First, multiply both sides by -1: \( -\sqrt{x + 5}=y + 3 \).
Then, subtract 3 from both sides: \( y=-\sqrt{x + 5}- 3 \)? Wait, no, wait. Wait, original function: \( y=(x + 3)^2 - 5 \), \( x < -3 \). Let's re - do the solving.
Starting from \( x=(y + 3)^2 - 5 \)
Add 5: \( x + 5=(y + 3)^2 \)
Take square roots: \( y + 3=\pm\sqrt{x + 5} \)
Now, the original function has \( x < -3 \), so \( x+3 < 0 \), and \( y=(x + 3)^2 - 5 \). When \( x < -3 \), as \( x\) decreases, \( (x + 3)^2 \) increases, so \( y\) increases? Wait, no, \( (x + 3)^2 \) is a parabola opening upwards with vertex at \( x=-3 \). For \( x < -3 \), the function \( (x + 3)^2 \) is decreasing (since the vertex is at \( x = - 3\) and for \( x < -3\), moving left from \( x=-3\), the value of \( (x + 3)^2\) decreases? Wait, no, if \( x=-4\), \( (x + 3)^2=(-1)^2 = 1\); if \( x=-5\), \( (x + 3)^2=(-2)^2 = 4\). Wait, that's increasing. So when \( x < -3\), \( (x + 3)^2\) is increasing (because as \( x\) becomes more negative, \( x + 3\) becomes more negative, and squaring a more negative number gives a larger positive number). So the original function \( y=(x + 3)^2-5\) for \( x < -3\) is increasing. So when we find the inverse, we should take the negative square root? Wait, no, let's think about the domain and range.
Original function: \( f(x)=(x + 3)^2-5\), \( x < -3\). The range of \( f(x)\): when \( x < -3\), \( (x + 3)^2>0\) (since \( x eq - 3\)), so \( (x + 3)^2-5>-5\). So the range of \( f(x)\) is \( y > - 5\). So the domain of the inverse function \( f^{-1}(x)\) is \( x > - 5\).
Now, going back to \( x=(y + 3)^2-5\), solve for \( y\):
\( (y + 3)^2=x + 5\)
Since the original function is increasing (for \( x < -3\)), the inverse function should be such that when we solve for \( y\), we take the negative square root? Wait, no, let's do the algebra again.
\( (y + 3)^2=x + 5\)
Take square roots: \( y + 3=\pm\sqrt{x + 5}\)
Now, in the original function, \( x < -3\), so \( y + 3=x + 3\) (wait, no, \( y=(x + 3)^2-5\), so \( y + 5=(x + 3)^2\), so \( x + 3=-\sqrt{y + 5}\) (because \( x < -3\), so \( x + 3 < 0\)). So when we swap \( x\) and \( y\), we have \( y + 3=-\sqrt{x + 5}\) (because in the original function, \( x + 3\) is negative, so in the inverse, \( y + 3\) should be negative).
So \( y+3 =-\sqrt{x + 5}\)
Subtract 3 from both sides: \( y=-\sqrt{x + 5}-3\)? Wait, no, that doesn't seem right. Wait, let's check with an example. Let's take \( x=-4\) (which is less than -3). Then \( f(-4)=(-4 + 3)^2-5=(-1)^2-5=1 - 5=-4\). So the point \((-4,-4)\) is on the original function. Now, for the inverse function, when \( x=-4\), \( y\) should be -4. Let's plug \( x=-4\) into \( y =-\sqrt{x + 5}-3\): \( y=-\sqrt{-4 + 5}-3=-\sqrt{1}-3=-1 - 3=-4\). That works. Now, let's check the domain of the inverse. The range of the original function: since \( x < -3\), \( (x + 3)^2>0\), so \( (x + 3)^2-5>-5\). So the domain of the inverse function is \( x > - 5\).

Wait, but let's re - express the steps correctly.
1. Start with \( y=(x + 3)^2-5\), \( x < -3\)
2. Swap \( x\) and \( y\): \( x=(y + 3)^2-5\)
3. Solve for \( y\):
- Add 5: \( x + 5=(y + 3)^2\)
- Take square roots: \( y + 3=\pm\sqrt{x + 5}\)
- Since \( x < -3\) in the original function, \( y + 3=x + 3<0\) (because \( x < -3\) implies \( x + 3 < 0\)). So when we solve for \( y\), we need \( y + 3<0\), so we take the negative square root: \( y + 3=-\sqrt{x + 5}\)
- Subtract 3: \( y=-\sqrt{x + 5}-3\)? Wait, no, that would be \( y=- \sqrt{x + 5}-3\), but let's check the options. Wait, maybe I made a mistake in the sign. Wait, the original function is \( f(x)=(x + 3)^2-5\), \( x < -3\). Let's consider the graph. The graph of \( y=(x + 3)^2-5\) is a parabola opening upwards with vertex at \((-3,-5)\). For \( x < -3\), the function is decreasing (wait, no, when \( x\) decreases (becomes more negative) from -3, \( (x + 3)^2\) increases, so the function is increasing for \( x < -3\)? Wait, when \( x=-4\), \( f(-4)=(-1)^2-5=-4\); when \( x=-5\), \( f(-5)=(-2)^2-5=-1\). So as \( x\) decreases (from -4 to -5), \( f(x)\) increases (from -4 to -1). So the function is increasing on \( x < -3\). So the inverse function should be a function that is also increasing? Wait, no, the inverse of an increasing function is increasing. Wait, maybe my earlier sign was wrong.
Wait, let's do the algebra again.
\( y=(x + 3)^2-5\)
\( x=(y + 3)^2-5\)
\( x + 5=(y + 3)^2\)
\( y + 3=\pm\sqrt{x + 5}\)
Since the original function is increasing on \( x < -3\), the inverse function should be increasing. So we need to take the negative square root? Wait, no, if the original function is increasing, the inverse should be increasing. Let's see, if we take \( y + 3=\sqrt{x + 5}\), then \( y=\sqrt{x + 5}-3\). Let's test with \( x=-4\): \( y=\sqrt{-4 + 5}-3=1 - 3=-2\), which is not equal to -4. If we take \( y + 3=-\sqrt{x + 5}\), then \( y=-\sqrt{x + 5}-3\), for \( x=-4\), \( y=-1 - 3=-4\), which is correct. But is the function increasing? Let's take \( x=-4\), \( y=-4\); \( x=-1\), \( y=-\sqrt{-1 + 5}-3=-\sqrt{4}-3=-2 - 3=-5\). Wait, as \( x\) increases from -4 to -1, \( y\) decreases from -4 to -5, so the inverse function is decreasing. But the original function is increasing (as \( x\) decreases from -3 to -5, \( f(x)\) increases from -5 to -1). So the inverse of an increasing function should be increasing. Wait, I think I messed up the direction of the original function. Let's find the derivative of \( f(x)=(x + 3)^2-5\), \( f^\prime(x)=2(x + 3)\). For \( x < -3\), \( f^\prime(x)<0\), so the original function is decreasing on \( x < -3\). Ah! That's my mistake. The derivative \( f^\prime(x) = 2(x + 3)\), when \( x < -3\), \( x+3 < 0\), so \( f^\prime(x)<0\), so the function is decreasing on \( x < -3\). So the inverse of a decreasing function is decreasing. So when we solve for \( y\), we need to take the negative square root to get a decreasing inverse.
So going back, \( y + 3=-\sqrt{x + 5}\), so \( y=-\sqrt{x + 5}-3\)? Wait, but let's check the domain of the inverse. The range of the original function: since \( x < -3\), \( (x + 3)^2>0\), so \( (x + 3)^2-5>-5\). So the domain of the inverse function is \( x > - 5\).

Wait, but let's check the options. The options for the inverse function: we have \( \sqrt{x + 5}\), and then operations. Wait, maybe I made a mistake in the sign. Wait, the original function is \( f(x)=(x + 3)^2-5\), \( x < -3\). Let's write \( x + 3=-\sqrt{y + 5}\) (because \( x < -3\), so \( x + 3 < 0\), and \( (x + 3)^2=y + 5\), so \( x + 3=-\sqrt{y + 5}\)). Then, solving for \( y\) in terms of \( x\) (after swapping \( x\) and \( y\)): \( y + 3=-\sqrt{x + 5}\), so \( y=-\sqrt{x + 5}-3\). But looking at the options, we have \( - \) (the minus sign), \( \sqrt{x + 5}\), and \( -3\) (which is \( +(-3)\)). Wait, the options have \( +3\), \( -3\), etc. Wait, maybe I made a mistake in the algebra. Let's start over.

Original function: \( y=(x + 3)^2-5\), \( x < -3\)

1. Replace \( f(x)\) with \( y\): \( y=(x + 3)^2-5\)
2. Swap \( x\) and \( y\): \( x=(y + 3)^2-5\)
3. Solve for \( y\):
- Add 5 to both sides: \( x + 5=(y + 3)^2\)
- Take the square root of both sides. Since \( x < -3\) in the original function, \( y + 3=x + 3<0\) (because \( y=(x + 3)^2-5\), so \( y + 5=(x + 3)^2\), so \( x + 3=-\sqrt{y + 5}\) as \( x < -3\)). So when we swap \( x\) and \( y\), we have \( y + 3=-\sqrt{x + 5}\)
- Subtract 3 from both sides: \( y=-\sqrt{x + 5}-3\)

Now, the domain of the inverse function: the range of the original function. Since \( x < -3\), \( (x + 3)^2>0\), so \( (x + 3)^2-5>-5\). So the domain of \( f^{-1}(x)\) is \( x > - 5\) (because the range of \( f(x)\) is \( y > - 5\), and the domain of the inverse is the range of the original function).

Now, looking at the options:
For the inverse function \( f^{-1}(x)\):
The expression: we have \( -\sqrt{x + 5}-3\), which can be written as \( - \sqrt{x + 5}+(-3)\). From the options, we have the minus sign, \( \sqrt{x + 5}\), and -3.
For the domain restriction: the range of the original function is \( y > - 5\), so the domain of the inverse is \( x > - 5\).

Answer:

The inverse function \( f^{-1}(x)=-\sqrt{x + 5}-3\) (which is composed of the minus sign, \( \sqrt{x + 5}\), and -3 from the options) and the domain restriction is \( x > - 5\) (using \( x > \) and -5 from the options). So filling in the blanks:
\( f^{-1}(x)=\boldsymbol{-\sqrt{x + 5}-3}\) (using the minus sign, \( \sqrt{x + 5}\), and -3) for \( \boldsymbol{x > - 5}\) (using \( x > \) and -5).