Find And Verify The Inverse Of F(x) = X^3 + 5

Hey math whizzes! Today, we're diving deep into the awesome world of inverse functions. Specifically, we're going to tackle a super cool problem involving the function $f(x) = x^3 + 5$. This function is a real gem because it's one-to-one, which means it has a unique inverse. Let's break down how to find that inverse and then prove it's the real deal. Get ready to flex those mathematical muscles, guys!

Part A: Finding the Inverse Function $f^{-1}(x)$

Alright, so our main mission here is to find the equation for $f^{-1}(x)$, the inverse function of $f(x) = x^3 + 5$. Remember, a function has an inverse if and only if it's one-to-one. This means that for every output value, there's only one input value. Think of it like a secret code – each message has only one original text. For $f(x) = x^3 + 5$, this property holds true. The cubic term, $x^3$, ensures that as $x$ increases, $f(x)$ also increases, and as $x$ decreases, $f(x)$ decreases. There are no turning points or loops that would cause multiple $x$ values to map to the same $y$ value. This is key to finding a unique inverse!

To find the inverse function, we follow a standard procedure. First, we replace $f(x)$ with $y$. So, we have $y = x^3 + 5$. The next crucial step is to swap $x$ and $y$. This is the magic move that essentially reverses the function's mapping. So, our equation becomes $x = y^3 + 5$. Now, our goal is to isolate $y$ in this new equation, because this isolated $y$ will be our inverse function, $f^{-1}(x)$.

Let's get solving! We have $x = y^3 + 5$. To start isolating $y^3$, we subtract 5 from both sides of the equation: $x - 5 = y^3$. Now, to get $y$ all by itself, we need to undo the cubing operation. The opposite of cubing a number is taking its cube root. So, we take the cube root of both sides: $\sqrt[3]{x - 5} = \sqrt[3]{y^3}$. This simplifies to $\sqrt[3]{x - 5} = y$. And there you have it! We've found our inverse function. We can now write it in the standard notation as $f^{-1}(x) = \sqrt[3]{x - 5}$.

Now, let's talk about the domain and range. For the original function $f(x) = x^3 + 5$, the domain is all real numbers (since you can cube any real number), and the range is also all real numbers. When we find an inverse function, the domain of the inverse function is the range of the original function, and the range of the inverse function is the domain of the original function. So, for $f^{-1}(x) = \sqrt[3]{x - 5}$, the domain is all real numbers, and the range is all real numbers. The cube root function itself is defined for all real numbers, so $x-5$ can be any real number, meaning $x$ can be any real number. There's no restriction on $x$ here, unlike in some other inverse functions involving square roots, for example. So, the condition 'x less lacksquare' you might see in some contexts for square root inverses isn't necessary here. The inverse function $f^{-1}(x) = \sqrt[3]{x - 5}$ is defined for all real numbers $x$.

So, to recap Part A, the equation for the inverse function is $f^{-1}(x) = \sqrt[3]{x - 5}$. Pretty straightforward once you remember to swap $x$ and $y$ and then solve for $y$. Keep this equation handy, because we're going to use it in the next part to verify our work!

Part B: Verifying the Inverse Function

Alright team, we've successfully found our candidate for the inverse function, $f^{-1}(x) = \sqrt[3]{x - 5}$. But in math, finding something is only half the battle; the other half is verifying that it's correct. We need to make sure that $f(x)$ and $f^{-1}(x)$ are truly inverse functions. The ultimate test for this is to show that when you compose them, you get the identity function, which is simply $x$. This means we need to prove two things: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. If both of these hold true, then we've nailed it!

Let's start with the first condition: $f(f^{-1}(x)) = x$. Remember, $f(x) = x^3 + 5$ and we found $f^{-1}(x) = \sqrt[3]{x - 5}$. To find $f(f^{-1}(x))$, we take our function $f$ and substitute the entire expression for $f^{-1}(x)$ wherever we see $x$ in the definition of $f$. So, in $f(x) = x^3 + 5$, we replace $x$ with $\sqrt[3]{x - 5}$. This gives us:

$f(f^{-1}(x)) = (f^{-1}(x))^3 + 5$ $f(f^{-1}(x)) = (\sqrt[3]{x - 5})^3 + 5$

Now, here's where the magic of inverse operations comes into play. The cube root and the cubing operation are opposites. They cancel each other out. So, $(\sqrt[3]{x - 5})^3$ simply simplifies to $(x - 5)$. Plugging this back into our equation, we get:

$f(f^{-1}(x)) = (x - 5) + 5$

And as you can see, the $-5$ and the $+5$ cancel each other out, leaving us with:

$f(f^{-1}(x)) = x$

Boom! The first condition is met. We've shown that composing $f$ with its inverse $f^{-1}$ in this order results in $x$. This is a huge step in confirming our inverse function is correct.

Now, let's move on to the second condition: $f^{-1}(f(x)) = x$. This is essentially doing the composition in the reverse order. We take our inverse function $f^{-1}(x) = \sqrt[3]{x - 5}$ and substitute the entire expression for $f(x)$ wherever we see $x$ in the definition of $f^{-1}$. So, in $f^{-1}(x) = \sqrt[3]{x - 5}$, we replace $x$ with $(x^3 + 5)$. This gives us:

$f^{-1}(f(x)) = \sqrt[3]{(f(x)) - 5}$ $f^{-1}(f(x)) = \sqrt[3]{(x^3 + 5) - 5}$

Again, we have some simplification happening inside the cube root. The $+5$ and the $-5$ cancel each other out, leaving us with:

$f^{-1}(f(x)) = \sqrt[3]{x^3}$

And just like before, the cube root and the cubing operation are inverse operations. They undo each other. So, $\sqrt[3]{x^3}$ simplifies beautifully to just $x$:

$f^{-1}(f(x)) = x$

And there you have it, guys! Both conditions, $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$, have been satisfied. This rigorous verification confirms that our equation for the inverse function, $f^{-1}(x) = \sqrt[3]{x - 5}$, is absolutely correct. It's always a good feeling to prove your work, right? This process is fundamental in understanding how functions and their inverses work together, creating a symmetrical relationship in the world of mathematics.

So, whether you're just starting with inverse functions or looking to solidify your understanding, remember these steps: find the inverse by swapping variables and solving, then verify by composing the functions in both orders. It's a powerful technique that works for many types of functions, though the algebra might get trickier with more complex ones. Keep practicing, and you'll be an inverse function master in no time! Happy solving!