Answer :
To find the relative extrema of the function [tex]\( g(x) = 3x^2(1 - x^2) \)[/tex], we will follow these steps:
### Step 1: Find the first derivative
First, we need to find the first derivative of [tex]\( g(x) \)[/tex].
[tex]\[ g(x) = 3x^2(1 - x^2) \][/tex]
We apply the product rule to differentiate:
[tex]\[ g'(x) = \frac{d}{dx}[3x^2(1 - x^2)] \][/tex]
[tex]\[ g'(x) = 3 \cdot 2x(1 - x^2) + 3x^2 \cdot (-2x) \][/tex]
[tex]\[ g'(x) = 6x(1 - x^2) - 6x^3 \][/tex]
[tex]\[ g'(x) = 6x - 6x^3 - 6x^3 \][/tex]
[tex]\[ g'(x) = 6x - 12x^3 \][/tex]
[tex]\[ g'(x) = 6x(1 - 2x^2) \][/tex]
### Step 2: Find the critical points
To find the critical points, set the first derivative equal to zero and solve for [tex]\( x \)[/tex].
[tex]\[ 6x(1 - 2x^2) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ 6x = 0 \quad \text{or} \quad 1 - 2x^2 = 0 \][/tex]
Solving these equations:
[tex]\[ x = 0 \][/tex]
and
[tex]\[ 1 - 2x^2 = 0 \][/tex]
[tex]\[ 2x^2 = 1 \][/tex]
[tex]\[ x^2 = \frac{1}{2} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{2}}{2} \][/tex]
So, the critical points are [tex]\( x = 0, \pm \frac{\sqrt{2}}{2} \)[/tex].
### Step 3: Find the second derivative
Next, find the second derivative of [tex]\( g(x) \)[/tex].
[tex]\[ g''(x) = \frac{d}{dx}[6x - 12x^3] \][/tex]
[tex]\[ g''(x) = 6 - 36x^2 \][/tex]
### Step 4: Apply the second derivative test
Evaluate the second derivative at each critical point to determine the nature of each extremum.
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ g''(0) = 6 - 36(0)^2 = 6 \][/tex]
Since [tex]\( g''(0) > 0 \)[/tex], this critical point is a relative minimum.
2. For [tex]\( x = \frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g''\left(\frac{\sqrt{2}}{2}\right) = 6 - 36\left(\frac{\sqrt{2}}{2}\right)^2 \][/tex]
[tex]\[ g''\left(\frac{\sqrt{2}}{2}\right) = 6 - 36\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g''\left(\frac{\sqrt{2}}{2}\right) = 6 - 18 = -12 \][/tex]
Since [tex]\( g''\left(\frac{\sqrt{2}}{2}\right) < 0 \)[/tex], this critical point is a relative maximum.
3. For [tex]\( x = -\frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g''\left(-\frac{\sqrt{2}}{2}\right) = 6 - 36\left(-\frac{\sqrt{2}}{2}\right)^2 \][/tex]
[tex]\[ g''\left(-\frac{\sqrt{2}}{2}\right) = 6 - 36\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g''\left(-\frac{\sqrt{2}}{2}\right) = 6 - 18 = -12 \][/tex]
Since [tex]\( g''\left(-\frac{\sqrt{2}}{2}\right) < 0 \)[/tex], this critical point is a relative maximum.
### Step 5: Find the function values at the critical points
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 3(0)^2(1 - (0)^2) = 0 \][/tex]
- For [tex]\( x = \frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3\left(\frac{\sqrt{2}}{2}\right)^2\left(1 - \left(\frac{\sqrt{2}}{2}\right)^2\right) \][/tex]
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(1 - \frac{1}{2}\right) \][/tex]
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3 \cdot \frac{1}{4} = \frac{3}{4} \][/tex]
- For [tex]\( x = -\frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3\left(-\frac{\sqrt{2}}{2}\right)^2\left(1 - \left(-\frac{\sqrt{2}}{2}\right)^2\right) \][/tex]
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(1 - \frac{1}{2}\right) \][/tex]
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3 \cdot \frac{1}{4} = \frac{3}{4} \][/tex]
### Summary of Results
- Relative maxima:
- [tex]\( \left( -\frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- [tex]\( \left( \frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- Relative minimum:
- [tex]\( \left(0, 0\right) \)[/tex]
Therefore, the answers to the question are:
- Relative maxima (smaller x-value) [tex]\((x, y)= \left( -\frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- [tex]\( (x, y) = \left( \frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- Relative minimum [tex]\((x, y)= (0, 0) \)[/tex]
### Step 1: Find the first derivative
First, we need to find the first derivative of [tex]\( g(x) \)[/tex].
[tex]\[ g(x) = 3x^2(1 - x^2) \][/tex]
We apply the product rule to differentiate:
[tex]\[ g'(x) = \frac{d}{dx}[3x^2(1 - x^2)] \][/tex]
[tex]\[ g'(x) = 3 \cdot 2x(1 - x^2) + 3x^2 \cdot (-2x) \][/tex]
[tex]\[ g'(x) = 6x(1 - x^2) - 6x^3 \][/tex]
[tex]\[ g'(x) = 6x - 6x^3 - 6x^3 \][/tex]
[tex]\[ g'(x) = 6x - 12x^3 \][/tex]
[tex]\[ g'(x) = 6x(1 - 2x^2) \][/tex]
### Step 2: Find the critical points
To find the critical points, set the first derivative equal to zero and solve for [tex]\( x \)[/tex].
[tex]\[ 6x(1 - 2x^2) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ 6x = 0 \quad \text{or} \quad 1 - 2x^2 = 0 \][/tex]
Solving these equations:
[tex]\[ x = 0 \][/tex]
and
[tex]\[ 1 - 2x^2 = 0 \][/tex]
[tex]\[ 2x^2 = 1 \][/tex]
[tex]\[ x^2 = \frac{1}{2} \][/tex]
[tex]\[ x = \pm \frac{\sqrt{2}}{2} \][/tex]
So, the critical points are [tex]\( x = 0, \pm \frac{\sqrt{2}}{2} \)[/tex].
### Step 3: Find the second derivative
Next, find the second derivative of [tex]\( g(x) \)[/tex].
[tex]\[ g''(x) = \frac{d}{dx}[6x - 12x^3] \][/tex]
[tex]\[ g''(x) = 6 - 36x^2 \][/tex]
### Step 4: Apply the second derivative test
Evaluate the second derivative at each critical point to determine the nature of each extremum.
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ g''(0) = 6 - 36(0)^2 = 6 \][/tex]
Since [tex]\( g''(0) > 0 \)[/tex], this critical point is a relative minimum.
2. For [tex]\( x = \frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g''\left(\frac{\sqrt{2}}{2}\right) = 6 - 36\left(\frac{\sqrt{2}}{2}\right)^2 \][/tex]
[tex]\[ g''\left(\frac{\sqrt{2}}{2}\right) = 6 - 36\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g''\left(\frac{\sqrt{2}}{2}\right) = 6 - 18 = -12 \][/tex]
Since [tex]\( g''\left(\frac{\sqrt{2}}{2}\right) < 0 \)[/tex], this critical point is a relative maximum.
3. For [tex]\( x = -\frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g''\left(-\frac{\sqrt{2}}{2}\right) = 6 - 36\left(-\frac{\sqrt{2}}{2}\right)^2 \][/tex]
[tex]\[ g''\left(-\frac{\sqrt{2}}{2}\right) = 6 - 36\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g''\left(-\frac{\sqrt{2}}{2}\right) = 6 - 18 = -12 \][/tex]
Since [tex]\( g''\left(-\frac{\sqrt{2}}{2}\right) < 0 \)[/tex], this critical point is a relative maximum.
### Step 5: Find the function values at the critical points
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 3(0)^2(1 - (0)^2) = 0 \][/tex]
- For [tex]\( x = \frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3\left(\frac{\sqrt{2}}{2}\right)^2\left(1 - \left(\frac{\sqrt{2}}{2}\right)^2\right) \][/tex]
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(1 - \frac{1}{2}\right) \][/tex]
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g\left(\frac{\sqrt{2}}{2}\right) = 3 \cdot \frac{1}{4} = \frac{3}{4} \][/tex]
- For [tex]\( x = -\frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3\left(-\frac{\sqrt{2}}{2}\right)^2\left(1 - \left(-\frac{\sqrt{2}}{2}\right)^2\right) \][/tex]
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(1 - \frac{1}{2}\right) \][/tex]
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \][/tex]
[tex]\[ g\left(-\frac{\sqrt{2}}{2}\right) = 3 \cdot \frac{1}{4} = \frac{3}{4} \][/tex]
### Summary of Results
- Relative maxima:
- [tex]\( \left( -\frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- [tex]\( \left( \frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- Relative minimum:
- [tex]\( \left(0, 0\right) \)[/tex]
Therefore, the answers to the question are:
- Relative maxima (smaller x-value) [tex]\((x, y)= \left( -\frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- [tex]\( (x, y) = \left( \frac{\sqrt{2}}{2}, \frac{3}{4} \right) \)[/tex]
- Relative minimum [tex]\((x, y)= (0, 0) \)[/tex]
