given f(x)=2x-15, g(x)=x²+19x and h(x)=√x+6, find (h°g)(-4)

To solve \( (h \circ g)(-4) \), where \( f(x) = 2x - 15 \), \( g(x) = x^2 + 19x \), and \( h(x) = \sqrt{x} + 6 \), we need to follow these steps:

### Step 1: Calculate \( g(-4) \)
First, evaluate \( g(x) \) at \( x = -4 \):
\[
g(-4) = (-4)^2 + 19(-4)
\]
\[
g(-4) = 16 - 76
\]
\[
g(-4) = -60
\]

### Step 2: Calculate \( h(g(-4)) = h(-60) \)
Next, evaluate \( h(x) \) at \( x = -60 \):
\[
h(-60) = \sqrt{-60} + 6
\]

However, the square root of a negative number \( \sqrt{-60} \) is not a real number, but rather a complex number. It can be expressed as:
\[
h(-60) = \sqrt{60}i + 6
\]

### Final Answer:
The expression \( (h \circ g)(-4) \) results in a complex number:
\[
(h \circ g)(-4) = \sqrt{60}i + 6
\]
where \( i \) represents the imaginary unit.

Would you like any further details, or do you have any questions?

### Related Questions:
1. What would the result be if \( h(x) \) was defined differently for negative inputs?
2. How would \( (h \circ g)(4) \) differ from \( (h \circ g)(-4) \)?
3. Can you explore the behavior of \( g(x) \) for different negative inputs?
4. What is the importance of considering the domain of a function before applying it?
5. How would the result change if \( g(x) \) were a linear function?
6. Can you explain how complex numbers are used in real-world applications?
7. What would happen if we applied \( f(x) \) after applying \( h(x) \)?
8. How does the composition of functions affect the overall domain of the resulting function?

### Tip:
Always check the domain and range of each function before performing compositions, especially when square roots or other operations that might lead to complex numbers are involved.

Explore how to calculate (h°g)(-4) where h(x) = √x + 6 and g(x) = x² + 19x. Understand the use of composition of functions and handling of square roots, resulting in a complex number solution. Suitable for advanced high school students.

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