Solve (h∘k)(x) if h(x)=3(2)^x and k(x)=3−2x

To solve \((h \circ k)(x)\), which represents the composition of functions \(h(x)\) and \(k(x)\), follow these steps:

1. **Given Functions**:
   - \(h(x) = 3 \cdot 2^x\)
   - \(k(x) = 3 - 2x\)

2. **Substitute \(k(x)\) into \(h(x)\)**:
   - \((h \circ k)(x) = h(k(x))\)
   - Replace \(x\) in \(h(x)\) with \(k(x)\):
   \[
   (h \circ k)(x) = 3 \cdot 2^{k(x)}
   \]

3. **Substitute the expression for \(k(x)\)**:
   \[
   (h \circ k)(x) = 3 \cdot 2^{3 - 2x}
   \]

So, the solution is:
\[
(h \circ k)(x) = 3 \cdot 2^{3 - 2x}
\]

Would you like a detailed explanation or have any questions?

Here are five related questions:
1. How do you find the inverse of a composite function?
2. What is the domain of the composite function \((h \circ k)(x)\)?
3. How would the solution change if \(h(x) = 2x + 3\) instead?
4. What does the graph of \((h \circ k)(x)\) look like?
5. How would you solve \((h \circ k)(x) = 0\)?

**Tip**: When working with composite functions, always substitute the inner function first before applying the outer function.

Learn how to solve the composite function (h∘k)(x) using the functions h(x)=3(2)^x and k(x)=3−2x. This problem involves understanding the composition of functions, exponential functions, and linear functions. Suitable for high school students.

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