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Question 1209854: For each of the following functions, determine if the function is increasing, decreasing, even, odd, and/or invertible on its natural domain. Select all the properties that apply.
f(x) = x^3 - 3x^2 + 10x - 15
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's analyze the function f(x) = x³ - 3x² + 10x - 15.
**1. Natural Domain:**
* Since f(x) is a polynomial, its natural domain is all real numbers, (-∞, ∞).
**2. Even or Odd:**
* **Even:** f(-x) = f(x)
* **Odd:** f(-x) = -f(x)
Let's test f(-x):
f(-x) = (-x)³ - 3(-x)² + 10(-x) - 15
f(-x) = -x³ - 3x² - 10x - 15
Now, let's see if f(-x) = f(x) or f(-x) = -f(x):
* f(x) = x³ - 3x² + 10x - 15
* -f(x) = -x³ + 3x² - 10x + 15
Comparing:
* f(-x) ≠ f(x) (Not even)
* f(-x) ≠ -f(x) (Not odd)
Therefore, the function is **neither even nor odd**.
**3. Increasing or Decreasing:**
* To determine if the function is increasing or decreasing, we need to analyze its derivative, f'(x).
f'(x) = 3x² - 6x + 10
To find when f'(x) > 0 (increasing) and when f'(x) < 0 (decreasing), we need to analyze the discriminant of the quadratic f'(x):
Discriminant (Δ) = b² - 4ac
Δ = (-6)² - 4(3)(10)
Δ = 36 - 120
Δ = -84
Since the discriminant is negative, the quadratic f'(x) has no real roots. Also, since the coefficient of x² in f'(x) is positive (3), the parabola opens upwards. This means f'(x) is always positive for all real numbers x.
Therefore, f'(x) > 0 for all x, and the function f(x) is **always increasing** on its domain.
**4. Invertibility:**
* If a function is strictly increasing or strictly decreasing, it is invertible.
* Since the function is strictly increasing over its domain, it is **invertible**.
**Conclusion:**
* **Domain:** (-∞, ∞)
* **Neither even nor odd**
* **Invertible:** Yes
* **Increasing:** Yes
* **Decreasing:** No
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