Questions: The function h is given by h(x)=x^4-6x^2 and the function k is given by k(x)=3x^2. Find all intervals where h(x)>k(x).
Transcript text: 20. The function $h$ is given by $h(x)=x^{4}-6 x^{2}$ and the function $k$ is given by $k(x)=3 x^{2}$. Find all intervals where $h(x)>k(x)$.
Solution
Solution Steps
Step 1: Set Up the Inequality
Given the functions \( h(x) = x^4 - 6x^2 \) and \( k(x) = 3x^2 \), we need to find the intervals where \( h(x) > k(x) \). This translates to solving the inequality:
\[
x^4 - 6x^2 > 3x^2
\]
Step 2: Simplify the Inequality
Simplify the inequality by combining like terms:
\[
x^4 - 9x^2 > 0
\]
Step 3: Factor the Inequality
Factor the expression on the left-hand side:
\[
x^2(x^2 - 9) > 0
\]
Further factorize:
\[
x^2(x - 3)(x + 3) > 0
\]
Step 4: Find the Critical Points
The critical points are the roots of the equation \( x^2(x - 3)(x + 3) = 0 \):
\[
x = 0, \, x = 3, \, x = -3
\]
Step 5: Test Intervals Between the Critical Points
Test the intervals determined by the critical points: \( (-\infty, -3) \), \( (-3, 0) \), \( (0, 3) \), and \( (3, \infty) \).
For \( x \in (-\infty, -3) \):
\[
x^2(x - 3)(x + 3) > 0 \quad \text{(True)}
\]
For \( x \in (-3, 0) \):
\[
x^2(x - 3)(x + 3) > 0 \quad \text{(False)}
\]
For \( x \in (0, 3) \):
\[
x^2(x - 3)(x + 3) > 0 \quad \text{(False)}
\]
For \( x \in (3, \infty) \):
\[
x^2(x - 3)(x + 3) > 0 \quad \text{(True)}
\]
Final Answer
The intervals where \( h(x) > k(x) \) are:
\[
\boxed{(-\infty, -3) \cup (3, \infty)}
\]