To find \((f-g-h)(x)\), we start by calculating \(f(x) - g(x) - h(x)\).
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Given functions:
- \(f(x) = -5x^2 + 4\)
- \(g(x) = -4x^2 - 7x + 9\)
- \(h(x) = 2x^2 + x - 2\)
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Calculating \(f - g\): \[ f(x) - g(x) = (-5x^2 + 4) - (-4x^2 - 7x + 9) \] Simplifying this: \[ = -5x^2 + 4 + 4x^2 + 7x - 9 \] \[ = (-5x^2 + 4x^2) + 7x + (4 - 9) \] \[ = -x^2 + 7x - 5 \]
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Now calculate \((f - g) - h\): \[ (f - g)(x) - h(x) = (-x^2 + 7x - 5) - (2x^2 + x - 2) \] Simplifying this: \[ = -x^2 + 7x - 5 - 2x^2 - x + 2 \] \[ = (-x^2 - 2x^2) + (7x - x) + (-5 + 2) \] \[ = -3x^2 + 6x - 3 \]
Thus, we find: \[ (f - g - h)(x) = -3x^2 + 6x - 3 \]
- Finding the domain: The functions \(f(x)\), \(g(x)\), and \(h(x)\) are all polynomials and thus are defined for all real numbers. Therefore, the domain of \((f - g - h)(x)\) is \((-\infty, \infty)\).
Final answer: \[ (f - g - h)(x) = -3x^2 + 6x - 3 \quad \text{and the domain of } (f - g - h)(x) \text{ is } (-\infty, \infty). \]
The correct option is: \((f-g-h)(x)=-3x^2+6x-3\) and the domain of \((f-g-h)(x)\) is \((-∞,∞)\).