What is the equation of the inverse of the function? f(x)=x^4+16, x ≥ 0

Mathematics

1 answer:

dangina [55]3 years ago

7 0

Answer:

https://tex.z-dn.net/?f=q%3D0%2C%5C%20r%3D2%2C%5C%20s%3D3%2C%5C%20t%3D-3

You might be interested in

Determine which equation is belongs to the graph of the limacon curve below.

JulsSmile [24]

Answer: Option C.

Step-by-step explanation:

We have functions of r(θ)

In our graph, we can see that the minimum value of r is when θ = 0°, and the maximum value is when θ = 180°.

We also know that the graph is in the square (-5, 5)x(-5, 5) and you can see that the maximum radius is almost less than 5.

Then let's analyze the options:

A) r = 3 - 2*cos(θ)

the maximum is at the right angle, but the maximum is:

r = 3 -2*(-1) = 5, so this maximum value is bigger than the one in the graph.

B) r = 3 - sin(θ)

For the sin functions, the maximum and minimum do not correspond with the values i write earlier, so we can discard this option.

C) r = 3 - cos(θ)

The maximum is: r = 3 - (-1) = 4, so this may be the correct answer.

the minimum is r = 3 - 1 = 2,

this is a possible equation for our circle.

D) r = 2 - 2*cos(θ)

Here, when θ = 0, we have: r = 2 - 2*1 = 0, but in the graph we can see that the radius is not 0 when θ = 0, so we can discard this option.

So the only option that makes sense is option C.

8 0

3 years ago

What can you say about his average speed?

Nikitich [7]

The answer should be E. 223- 6s = 0

5 0

3 years ago

Factor completely x^8- 16.

Nataly_w [17]

Answer:

Step-by-step explanation:

So in this example we'll be using the difference of squares which essentially states that: $(a-b)(a+b)=a^2-b^2$ or another way to think of it would be: $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

$x^8-16 = (x^4-4)(x^4+4)$

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

$x^4-4 = (x^2-2)(x^2+2)$

So completely factored form is: $(x^2-2)(x^2+4)(x^4+4)$

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because $x^2+4 = x^2-(-4)$ . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: $x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i)$ just something that might be useful in some cases.