We are given three functions: $f(x) = x^4 + 3$, $g(x) = x - 6$, and $h(x) = \sqrt{x}$. We need to find the composite function $f(g(h(x)))$.

AlgebraFunctionsComposite FunctionsFunction CompositionAlgebraic ManipulationExponents
2025/7/3

1. Problem Description

We are given three functions: f(x)=x4+3f(x) = x^4 + 3, g(x)=x6g(x) = x - 6, and h(x)=xh(x) = \sqrt{x}. We need to find the composite function f(g(h(x)))f(g(h(x))).

2. Solution Steps

First, we need to find g(h(x))g(h(x)). We substitute h(x)h(x) into g(x)g(x):
g(h(x))=h(x)6=x6g(h(x)) = h(x) - 6 = \sqrt{x} - 6
Next, we need to find f(g(h(x)))f(g(h(x))). We substitute g(h(x))g(h(x)) into f(x)f(x):
f(g(h(x)))=f(x6)=(x6)4+3f(g(h(x))) = f(\sqrt{x} - 6) = (\sqrt{x} - 6)^4 + 3
Expanding (x6)4(\sqrt{x} - 6)^4 can be cumbersome. Let's rewrite it. (x6)2=x12x+36(\sqrt{x} - 6)^2 = x - 12\sqrt{x} + 36. Then, (x6)4=(x12x+36)2=(x12x+36)(x12x+36)=x2+144x+129624xx+72x7212x+723672xx=x2+216x+1296+5184144xx864x=x2144xx+216x864x+6480(\sqrt{x} - 6)^4 = (x - 12\sqrt{x} + 36)^2 = (x - 12\sqrt{x} + 36)(x - 12\sqrt{x} + 36) = x^2 + 144x + 1296 - 24x\sqrt{x} + 72x - 72*12\sqrt{x} + 72*36 - 72x\sqrt{x} = x^2+216x + 1296+5184 - 144x\sqrt{x} -864\sqrt{x}=x^2 -144x\sqrt{x} +216x -864\sqrt{x}+6480.
So f(g(h(x)))=(x6)4+3=x2144xx+216x864x+6480+3=x2144xx+216x864x+6483f(g(h(x))) = (\sqrt{x} - 6)^4 + 3 = x^2 - 144x\sqrt{x} + 216x - 864\sqrt{x} + 6480+3 = x^2 - 144x\sqrt{x} + 216x - 864\sqrt{x} + 6483.
However, since we're inputting into myopenmath, they would probably like the expression (x6)4+3(\sqrt{x}-6)^4+3.

3. Final Answer

(x6)4+3(\sqrt{x} - 6)^4 + 3