To find \((f−g−h)(x)\), we will start by defining the functions \(f(x)\), \(g(x)\), and \(h(x)\):
- \(f(x) = 5x^2 + 3x - 6\)
- \(g(x) = |x + 1| - 7\)
- \(h(x) = -x^4 + 2x^2 + 3\)
The expression for \((f - g - h)(x)\) is given by: \[ (f - g - h)(x) = f(x) - g(x) - h(x) \]
Now, we can substitute the functions in:
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Start with \(f(x)\): \[ f(x) = 5x^2 + 3x - 6 \]
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Subtract \(g(x)\): \[ g(x) = |x + 1| - 7 \] So, \[ f(x) - g(x) = (5x^2 + 3x - 6) - (|x + 1| - 7) = 5x^2 + 3x - 6 - |x + 1| + 7 \] Which simplifies to: \[ f(x) - g(x) = 5x^2 + 3x + 1 - |x + 1| \]
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Now subtract \(h(x)\): \[ h(x) = -x^4 + 2x^2 + 3 \] So, \[ (f - g)(x) - h(x) = (5x^2 + 3x + 1 - |x + 1|) - (-x^4 + 2x^2 + 3) \] That becomes: \[ f(x) - g(x) - h(x) = 5x^2 + 3x + 1 - |x + 1| + x^4 - 2x^2 - 3 \] Simplifying further: \[ = x^4 + (5x^2 - 2x^2) + 3x + (1 - 3) - |x + 1| \] \[ = x^4 + 3x^2 + 3x - |x + 1| - 2 \]
Thus, we have: \[ (f - g - h)(x) = x^4 + 3x^2 + 3x - |x + 1| - 2 \]
Now let’s find the domain of \((f - g - h)(x)\):
- Since \(f(x)\), \(g(x)\), and \(h(x)\) are all polynomials or piecewise continuous functions, their domains are all real numbers.
- The absolute value function \(g(x) = |x + 1| - 7\) does not impose any restrictions as it is defined for all \(x\).
Thus, the domain of \((f - g - h)(x)\) is:
\[ (-\infty, \infty) \]
Final Answer:
\((f - g - h)(x) = x^4 + 3x^2 + 3x - |x + 1| - 2\) and the domain of \((f - g - h)(x)\) is \((-∞, ∞)\).