Answer :
To determine the graph of the function \( f(x) = 0.03 x^2(x^2 - 25) \), let's go through a detailed, step-by-step analysis.
1. Simplifying the Function:
First, we simplify the function:
[tex]\[ f(x) = 0.03 x^2 (x^2 - 25) = 0.03 (x^4 - 25x^2) \][/tex]
So,
[tex]\[ f(x) = 0.03 x^4 - 0.75 x^2 \][/tex]
2. Finding Critical Points:
To find the critical points, we need to find where the first derivative, \( f'(x) \), is equal to zero.
[tex]\[ f'(x) = 0.12 x^3 - 1.5 x \][/tex]
Setting \( f'(x) = 0 \):
[tex]\[ 0.12 x^3 - 1.5 x = 0 \][/tex]
Factor out \( x \):
[tex]\[ x (0.12 x^2 - 1.5) = 0 \][/tex]
Setting each factor to zero gives us:
[tex]\[ x = 0 \quad \text{or} \quad 0.12 x^2 - 1.5 = 0 \][/tex]
Solving \( 0.12 x^2 - 1.5 = 0 \):
[tex]\[ 0.12 x^2 = 1.5 \][/tex]
[tex]\[ x^2 = \frac{1.5}{0.12} = 12.5 \][/tex]
[tex]\[ x = \pm \sqrt{12.5} = \pm \sqrt{25 \cdot 0.5} = \pm 5\sqrt{0.5} \approx \pm 3.536 \][/tex]
Therefore, the critical points are:
[tex]\[ x = -3.536, 0, 3.536 \][/tex]
3. Second Derivative Test:
To determine the concavity at these critical points, we evaluate the second derivative \( f''(x) \):
[tex]\[ f''(x) = 0.36 x^2 - 1.5 \][/tex]
Evaluating \( f''(x) \) at the critical points:
- At \( x = -3.536 \):
[tex]\[ f''(-3.536) = 0.36 (-3.536)^2 - 1.5 \approx 3 \][/tex]
- At \( x = 0 \):
[tex]\[ f''(0) = 0.36 (0)^2 - 1.5 = -1.5 \][/tex]
- At \( x = 3.536 \):
[tex]\[ f''(3.536) = 0.36 (3.536)^2 - 1.5 \approx 3 \][/tex]
So, the concavities are:
[tex]\[ f''(-3.536) \approx 3 \quad (\text{concave up}) \][/tex]
[tex]\[ f''(0) = -1.5 \quad (\text{concave down}) \][/tex]
[tex]\[ f''(3.536) \approx 3 \quad (\text{concave up}) \][/tex]
4. Function Values at Critical Points:
We now find the function values at specific points including the critical points:
- At \( x = 0 \):
[tex]\[ f(0) = 0.03 (0)^4 - 0.75 (0)^2 = 0 \][/tex]
- At \( x = 5 \):
[tex]\[ f(5) = 0.03 (5)^4 - 0.75 (5)^2 = 0.03 \cdot 625 - 0.75 \cdot 25 = 18.75 - 18.75 = 0 \][/tex]
- At \( x = -5 \):
[tex]\[ f(-5) = 0.03 (-5)^4 - 0.75 (-5)^2 = 0.03 \cdot 625 - 0.75 \cdot 25 = 18.75 - 18.75 = 0 \][/tex]
5. Signature of the Graph:
From the above analysis, we can conclude the following:
- The function \( f(x) = 0.03 x^4 - 0.75 x^2 \) has zeros at \( x = 0, x = 5, \) and \( x = -5 \).
- The critical points are at \( x \approx -3.536, 0, 3.536 \).
- The function is concave up at \( x \approx \pm 3.536 \) and concave down at \( x = 0 \).
This gives the signature of a quartic polynomial with specific symmetry and concavity properties.
Based on these details, we should look for a graph that passes through the origin and the points [tex]\( x = \pm 5 \)[/tex], has local minima near [tex]\( x \approx \pm 3.536 \)[/tex], and shows a local maximum at the origin [tex]\( x = 0 \)[/tex].
1. Simplifying the Function:
First, we simplify the function:
[tex]\[ f(x) = 0.03 x^2 (x^2 - 25) = 0.03 (x^4 - 25x^2) \][/tex]
So,
[tex]\[ f(x) = 0.03 x^4 - 0.75 x^2 \][/tex]
2. Finding Critical Points:
To find the critical points, we need to find where the first derivative, \( f'(x) \), is equal to zero.
[tex]\[ f'(x) = 0.12 x^3 - 1.5 x \][/tex]
Setting \( f'(x) = 0 \):
[tex]\[ 0.12 x^3 - 1.5 x = 0 \][/tex]
Factor out \( x \):
[tex]\[ x (0.12 x^2 - 1.5) = 0 \][/tex]
Setting each factor to zero gives us:
[tex]\[ x = 0 \quad \text{or} \quad 0.12 x^2 - 1.5 = 0 \][/tex]
Solving \( 0.12 x^2 - 1.5 = 0 \):
[tex]\[ 0.12 x^2 = 1.5 \][/tex]
[tex]\[ x^2 = \frac{1.5}{0.12} = 12.5 \][/tex]
[tex]\[ x = \pm \sqrt{12.5} = \pm \sqrt{25 \cdot 0.5} = \pm 5\sqrt{0.5} \approx \pm 3.536 \][/tex]
Therefore, the critical points are:
[tex]\[ x = -3.536, 0, 3.536 \][/tex]
3. Second Derivative Test:
To determine the concavity at these critical points, we evaluate the second derivative \( f''(x) \):
[tex]\[ f''(x) = 0.36 x^2 - 1.5 \][/tex]
Evaluating \( f''(x) \) at the critical points:
- At \( x = -3.536 \):
[tex]\[ f''(-3.536) = 0.36 (-3.536)^2 - 1.5 \approx 3 \][/tex]
- At \( x = 0 \):
[tex]\[ f''(0) = 0.36 (0)^2 - 1.5 = -1.5 \][/tex]
- At \( x = 3.536 \):
[tex]\[ f''(3.536) = 0.36 (3.536)^2 - 1.5 \approx 3 \][/tex]
So, the concavities are:
[tex]\[ f''(-3.536) \approx 3 \quad (\text{concave up}) \][/tex]
[tex]\[ f''(0) = -1.5 \quad (\text{concave down}) \][/tex]
[tex]\[ f''(3.536) \approx 3 \quad (\text{concave up}) \][/tex]
4. Function Values at Critical Points:
We now find the function values at specific points including the critical points:
- At \( x = 0 \):
[tex]\[ f(0) = 0.03 (0)^4 - 0.75 (0)^2 = 0 \][/tex]
- At \( x = 5 \):
[tex]\[ f(5) = 0.03 (5)^4 - 0.75 (5)^2 = 0.03 \cdot 625 - 0.75 \cdot 25 = 18.75 - 18.75 = 0 \][/tex]
- At \( x = -5 \):
[tex]\[ f(-5) = 0.03 (-5)^4 - 0.75 (-5)^2 = 0.03 \cdot 625 - 0.75 \cdot 25 = 18.75 - 18.75 = 0 \][/tex]
5. Signature of the Graph:
From the above analysis, we can conclude the following:
- The function \( f(x) = 0.03 x^4 - 0.75 x^2 \) has zeros at \( x = 0, x = 5, \) and \( x = -5 \).
- The critical points are at \( x \approx -3.536, 0, 3.536 \).
- The function is concave up at \( x \approx \pm 3.536 \) and concave down at \( x = 0 \).
This gives the signature of a quartic polynomial with specific symmetry and concavity properties.
Based on these details, we should look for a graph that passes through the origin and the points [tex]\( x = \pm 5 \)[/tex], has local minima near [tex]\( x \approx \pm 3.536 \)[/tex], and shows a local maximum at the origin [tex]\( x = 0 \)[/tex].
