Let f(x)= x^3, g(x) = sqrt x, h(x) = x-4, and j(x)= 2x. Express the following function k as a composite of three of these four functions.
k(x)= sqrt (x^3 - 4)
Do I need to factor anything out of the parentheses or something? If you can, please explain this problem to me :) any help is GREATLY appreciated!!
this didn’t help me at all. y’all suck .
Ohhh that makes perfect sense! I don't see how I couldn't see that before. Soooo the answer would be g(h(f(x))), right?
Thanks for your help!! :D :D
It’s 2 square 2
To express the function k(x) as a composite of three of the given functions, we need to break down the expression sqrt(x^3 - 4) and find the combination of functions that can produce it.
Let's begin by examining the expression sqrt(x^3 - 4).
Notice that the expression inside the square root symbol (x^3 - 4) involves raising x to the power of 3 and subtracting 4.
Among the given functions, the function f(x) = x^3 is the closest one that involves raising x to the power of 3. So, let's start by using f(x): k(x) = sqrt(f(x) - 4).
Now, we need to eliminate the constant term (-4) in the expression inside the square root. The function h(x) = x - 4 subtracts 4 from x, so we can use this function to achieve our goal: k(x) = sqrt(f(x) - h(x)).
Finally, we have a remaining square root operation. From the given functions, the function g(x) = sqrt(x) handles square root operations. Therefore, we can express the function k(x) as: k(x) = g(f(x) - h(x)).
So, the function k(x) can be written as a composite of the three functions: k(x) = g(f(x) - h(x)).
If you want to simplify this further, you can substitute the definitions of f(x), g(x), and h(x) into the equation and perform the necessary calculations.
Glad to be of help!
The composition of functions can be expressed as the function of a function.
Let f(x)=x² and g(x) = x+1,
fog(x) = f(g(x)) = f(x+1)=(x+1)²
gof(x) = g(f(x)) = g(x²) = x²+1
So it is clear that in general fog(x) ≠ gof(x).
You can read more about it in this article:
http://en.wikipedia.org/wiki/Function_composition
Given f(x)= x^3, g(x) = sqrt x, h(x) = x-4, and j(x)= 2x.
You can "compose" the function
m(x) = x³-4 from
m(x)
= hof(x)
= h(f(x))
= h(x³)
= x³-4
Can you take it from here?