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To solve the inequality [tex]\( g(x) = x^5 - 4x^3 > 0 \)[/tex], let us follow these steps:
1. Factor the function:
First, we factor [tex]\( g(x) \)[/tex]. Notice that you can factor out [tex]\( x^3 \)[/tex] from the expression:
[tex]\[ g(x) = x^5 - 4x^3 = x^3(x^2 - 4). \][/tex]
Further, we can factor [tex]\( x^2 - 4 \)[/tex] as the difference of squares:
[tex]\[ x^2 - 4 = (x - 2)(x + 2). \][/tex]
Thus, the factored form of [tex]\( g(x) \)[/tex] is:
[tex]\[ g(x) = x^3 (x - 2)(x + 2). \][/tex]
2. Find the critical points:
Critical points are the values of [tex]\( x \)[/tex] where [tex]\( g(x) = 0 \)[/tex]. Setting the factored form equal to zero, we get:
[tex]\[ x^3(x - 2)(x + 2) = 0. \][/tex]
This gives us critical points at:
[tex]\[ x = 0, \quad x = 2, \quad x = -2. \][/tex]
3. Determine intervals:
The critical points divide the real number line into several intervals. We need to test the sign of [tex]\( g(x) \)[/tex] in each interval. The intervals are:
[tex]\[ (-\infty, -2), \quad (-2, 0), \quad (0, 2), \quad (2, \infty). \][/tex]
4. Test the sign of [tex]\( g(x) \)[/tex] in each interval:
We can pick a test point [tex]\( x \)[/tex] from each interval and evaluate the sign of [tex]\( g(x) \)[/tex].
- For [tex]\( x \in (-\infty, -2) \)[/tex]:
Choose [tex]\( x = -3 \)[/tex]:
[tex]\[ g(-3) = (-3)^3 (-3 - 2)(-3 + 2) = -27 \cdot -5 \cdot -1 = -135 < 0. \][/tex]
So, [tex]\( g(x) < 0 \)[/tex] in [tex]\( (-\infty, -2) \)[/tex].
- For [tex]\( x \in (-2, 0) \)[/tex]:
Choose [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = (-1)^3 (-1 - 2)(-1 + 2) = -1 \cdot -3 \cdot 1 = 3 > 0. \][/tex]
So, [tex]\( g(x) > 0 \)[/tex] in [tex]\( (-2, 0) \)[/tex].
- For [tex]\( x \in (0, 2) \)[/tex]:
Choose [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = 1^3 (1 - 2)(1 + 2) = 1 \cdot -1 \cdot 3 = -3 < 0. \][/tex]
So, [tex]\( g(x) < 0 \)[/tex] in [tex]\( (0, 2) \)[/tex].
- For [tex]\( x \in (2, \infty) \)[/tex]:
Choose [tex]\( x = 3 \)[/tex]:
[tex]\[ g(3) = 3^3 (3 - 2)(3 + 2) = 27 \cdot 1 \cdot 5 = 135 > 0. \][/tex]
So, [tex]\( g(x) > 0 \)[/tex] in [tex]\( (2, \infty) \)[/tex].
5. Combine the intervals:
We are looking for where [tex]\( g(x) > 0 \)[/tex]. From the intervals tested, [tex]\( g(x) \)[/tex] is positive in [tex]\( (-2, 0) \)[/tex] and [tex]\( (2, \infty) \)[/tex].
Therefore, the solution to the inequality [tex]\( g(x) > 0 \)[/tex] is:
[tex]\[ (-2, 0) \cup (2, \infty). \][/tex]
1. Factor the function:
First, we factor [tex]\( g(x) \)[/tex]. Notice that you can factor out [tex]\( x^3 \)[/tex] from the expression:
[tex]\[ g(x) = x^5 - 4x^3 = x^3(x^2 - 4). \][/tex]
Further, we can factor [tex]\( x^2 - 4 \)[/tex] as the difference of squares:
[tex]\[ x^2 - 4 = (x - 2)(x + 2). \][/tex]
Thus, the factored form of [tex]\( g(x) \)[/tex] is:
[tex]\[ g(x) = x^3 (x - 2)(x + 2). \][/tex]
2. Find the critical points:
Critical points are the values of [tex]\( x \)[/tex] where [tex]\( g(x) = 0 \)[/tex]. Setting the factored form equal to zero, we get:
[tex]\[ x^3(x - 2)(x + 2) = 0. \][/tex]
This gives us critical points at:
[tex]\[ x = 0, \quad x = 2, \quad x = -2. \][/tex]
3. Determine intervals:
The critical points divide the real number line into several intervals. We need to test the sign of [tex]\( g(x) \)[/tex] in each interval. The intervals are:
[tex]\[ (-\infty, -2), \quad (-2, 0), \quad (0, 2), \quad (2, \infty). \][/tex]
4. Test the sign of [tex]\( g(x) \)[/tex] in each interval:
We can pick a test point [tex]\( x \)[/tex] from each interval and evaluate the sign of [tex]\( g(x) \)[/tex].
- For [tex]\( x \in (-\infty, -2) \)[/tex]:
Choose [tex]\( x = -3 \)[/tex]:
[tex]\[ g(-3) = (-3)^3 (-3 - 2)(-3 + 2) = -27 \cdot -5 \cdot -1 = -135 < 0. \][/tex]
So, [tex]\( g(x) < 0 \)[/tex] in [tex]\( (-\infty, -2) \)[/tex].
- For [tex]\( x \in (-2, 0) \)[/tex]:
Choose [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = (-1)^3 (-1 - 2)(-1 + 2) = -1 \cdot -3 \cdot 1 = 3 > 0. \][/tex]
So, [tex]\( g(x) > 0 \)[/tex] in [tex]\( (-2, 0) \)[/tex].
- For [tex]\( x \in (0, 2) \)[/tex]:
Choose [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = 1^3 (1 - 2)(1 + 2) = 1 \cdot -1 \cdot 3 = -3 < 0. \][/tex]
So, [tex]\( g(x) < 0 \)[/tex] in [tex]\( (0, 2) \)[/tex].
- For [tex]\( x \in (2, \infty) \)[/tex]:
Choose [tex]\( x = 3 \)[/tex]:
[tex]\[ g(3) = 3^3 (3 - 2)(3 + 2) = 27 \cdot 1 \cdot 5 = 135 > 0. \][/tex]
So, [tex]\( g(x) > 0 \)[/tex] in [tex]\( (2, \infty) \)[/tex].
5. Combine the intervals:
We are looking for where [tex]\( g(x) > 0 \)[/tex]. From the intervals tested, [tex]\( g(x) \)[/tex] is positive in [tex]\( (-2, 0) \)[/tex] and [tex]\( (2, \infty) \)[/tex].
Therefore, the solution to the inequality [tex]\( g(x) > 0 \)[/tex] is:
[tex]\[ (-2, 0) \cup (2, \infty). \][/tex]
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