Im very confused with these math problems if someone can help

Question

Consider the following function.

f(x) = x⁴ + 2,    x ≥ 0

Find the inverse function f ⁻¹.

f ⁻¹(x) = 

State the domain and range of f.

Domain:

Range:

State the domain and range of f ⁻¹.

 (Enter your answers using interval notation.)

Domain :

Range:

Answer 1

They give you the domain of the function, x ≥ 0. In interval notation, you write that the domain of f(x) is [0, ∞).

A good way to see the range is to sketch the function. To do that, plug three values for x ≥ 0 into f(x) = x⁴ + 2, say f(0) = 0 + 2 = 2, f(1) = 1 + 2 = 3, f(2) = 16 + 2 = 18, and mark those points (0, 2),(1, 3),(2, 18) on an x-y plane and sketch a curve through them. Because the domain is [0, ∞), your sketch should not go where x < 0. It’s only a half of a parabola. Now, you can see that the range is f(x) ≥ 2, which is [2, ∞) in interval notation.

The cool part about inverses is that they are pretty easy to draw. The inverse is just the mirror image of the original graph across the line y = x, which is the upward diagonal line going through the origin (0, 0). (You can draw this as a dotted line so you see it.) To sketch the inverse function, first switch the x and y of the three points to get (2, 0),(3, 1),(18, 2), and mark those points on the x-y plane. Then, sketch the curve through these points so it’s a nice mirror image of the original graph across the dotted line. It’s the half of a parabola turned sideways.

You can see the domain and range of the inverse. The domain, on the x-axis, is [2, ∞). And, because it’s only the half of the parabola, the range is [0, ∞).

Now that we know intuitively what the function and its inverse look like, we can figure out what the inverse f⁻¹ is using algebra. A good insight is that the function’s inverse ”undoes” the function. If you plug the output of the function f(x) into the inverse, you get the original x back. That is, f⁻¹(f(x)) = x, It also works the other way around: f(f⁻¹(x)) = x.

Let’s use this concept to figure out what f⁻¹ (x) is. First, let’s make y = f⁻¹ (x) so it’s easier to see, so f(f⁻¹(x)) = x would be written f(y) = x or y⁴ + 2 = x. Doing some algebra we get y⁴ = x − 2. which gives y = ± the fourth root of x − 2. So the inverse is f⁻¹(x) = ± fourth root of x − 2.

This plus or minus is the confusing part! If we plug in our inverse points from above: For example, (18, 2), we get f⁻¹(18) = ± fourth root of (18 − 2) = ±2. The algebra answer gives us both the positive and negative numbers, but the graph only has the positive numbers. What is going on?

Here’s the answer. Because the original function has a restricted domain, in this case only x ≥ 0, then the range of the inverse function is also restricted to those numbers. So, while −1 is one possible answer of f⁻¹(3), there is no place for that answer to go, so you just drop it.

You’ll see this kind of situation in physics. When throwing a ball, the equation for how long the ball stays in the air could look something like 0 = −10t² + 5t + 2, where t is the number of seconds starting when you throw it. Using the quadratic formula, you will get an answer with a positive t and a negative t. But the domain of t is t ≥ 0. You can’t go backwards in time! So you drop the negative t.

Answer 2

The answer is "almost" there is no inverse Function. DNE. Does Not Exist. Use a graphing calulator to graph it. It's very similar to a parabola. vertical parabolas dont' have inverse functions. the inverse fails the vertical line test.

but x is restricted to non negative numbers,so it's really just half of graph, which does have an inverse

y=x^4 +2

switch x and y, then solve for the new y to get the inverse relation

x=y^4+2

y^4=x-2

y=f^-1(x)= (x-2)^.25 = the 4th root of x-2 = the square root of the square root of x-2

for f(x)

domain would be any real number, but the problem expressly says greater than or equal to 0

domain: x>0

range x>2

for f^-1, switch them, just the reverse of f(x)

domain: x>2

^{range x>0}