Let `f(x) = x^2, g(x) = sin x, h(x) =sqrtx,` then verify that `[fo(goh)] (x)` and `[(fog)oh] (x)` are equal.

Explanation

To verify that \([f \circ (g \circ h)](x)\) and \([(f \circ g) \circ h](x)\) are equal, we will follow these steps:

### Step 1: Define the Functions
We have:
- \( f(x) = x^2 \)
- \( g(x) = \sin x \)
- \( h(x) = \sqrt{x} \)

### Step 2: Calculate \( g \circ h \)
We need to find \( g(h(x)) \):
\[ g(h(x)) = g(\sqrt{x}) = \sin(\sqrt{x}) \]

### Step 3: Calculate \( f \circ (g \circ h) \)
Now we need to find \( f(g(h(x))) \):
\[ f(g(h(x))) = f(\sin(\sqrt{x})) = (\sin(\sqrt{x}))^2 \]

### Step 4: Calculate \( f \circ g \)
Next, we find \( f(g(x)) \):
\[ f(g(x)) = f(\sin x) = (\sin x)^2 \]

### Step 5: Calculate \((f \circ g) \circ h\)
Now we need to find \((f \circ g)(h(x))\):
\[ (f \circ g)(h(x)) = f(g(h(x))) = f(\sin(\sqrt{x})) = (\sin(\sqrt{x}))^2 \]

### Step 6: Compare the Results
Now we compare the results from Step 3 and Step 5:
\[ f(g(h(x))) = (\sin(\sqrt{x}))^2 \]
\[ (f \circ g)(h(x)) = (\sin(\sqrt{x}))^2 \]

Since both expressions are equal, we conclude that:
\[ [f \circ (g \circ h)](x) = [(f \circ g) \circ h](x) \]

Thus, we have verified that \([f \circ (g \circ h)](x) = [(f \circ g) \circ h](x)\).

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