turns-00028.parquet:62833
70cc8d321f489e00917edfc0degenerate_repetitionAbsentFinal dense release
USER
The intersection is obtained by setting in the equation for y=0, which gives
z=3x^{2}\cdot e^{-x^{2}}
To analyze what this function looks like, we note that it tends towards for very large positive or negative values of . We need to understand how many local maxima and minima it has to better understand its shape.
One way to do this is to note that it is 0 at 0, it is always positive. (In particular it is an even function, meaning it is symmetric about .) Therefore it has one minimum at 0, and must have 2 maxima.
Another way to analyze the shape is to take the derivative, identify critical points, and then analyze the critical points directly using the first or second derivative test from single variable calculus.ASSISTANT
You're on the right track! Let's carry out the analysis step by step to determine the number and nature of the local maxima and minima of the function:
\[
z(x) = 3x^{2}e^{-x^{2}}
\]
### Step 1: Find the Critical Points
First, we need to find the critical points by taking the first derivative of \( z(x) \) with respect to \( x \) and setting it equal to zero.
**Compute the first derivative \( \frac{dz}{dx} \):**
We'll use the product rule for differentiation, which states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then:
\[
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
\]
Let:
- \( u(x) = 3x^{2} \)
- \( v(x) = e^{-x^{2}} \)
Compute \( u'(x) \) and \( v'(x) \):
\[
u'(x) = \frac{d}{dx}[3x^{2}] = 6x
\]
\[
v'(x) = \frac{d}{dx}[e^{-x^{2}}] = e^{-x^{2}} \cdot (-2x) = -2xe^{-x^{2}}
\]
Now, apply the product rule:
\[
\frac{dz}{dx} = u'(x)v(x) + u(x)v'(x) = [6x \cdot e^{-x^{2}}] + [3x^{2} \cdot (-2xe^{-x^{2}})]
\]
Simplify:
\[
\frac{dz}{dx} = 6xe^{-x^{2}} - 6x^{3}e^{-x^{2}}
\]
Factor out common terms:
\[
\frac{dz}{dx} = 6xe^{-x^{2}}(1 - x^{2})
\]
**Set the first derivative equal to zero:**
\[
6xe^{-x^{2}}(1 - x^{2}) = 0
\]
Since \( e^{-x^{2}} > 0 \) for all real \( x \), we can ignore this factor when solving for critical points. So, set the remaining factors to zero:
1. \( x = 0 \)
2. \( 1 - x^{2} = 0 \) ⟹ \( x^{2} = 1 \) ⟹ \( x = \pm 1 \)
**Critical Points:** \( x = -1, 0, 1 \)
---
### Step 2: Determine the Nature of Critical Points
We'll use the **second derivative test** to classify each critical point.
**Compute the second derivative \( \frac{d^{2}z}{dx^{2}} \):**
Start from the simplified form of the first derivative:
\[
\frac{dz}{dx} = 6xe^{-x^{2}}(1 - x^{2})
\]
We'll differentiate this using the product rule and chain rule.
Let:
\[
w(x) = 6xe^{-x^{2}}, \quad y(x) = (1 - x^{2})
\]
Compute \( w'(x) \):
First, \( w(x) = 6xe^{-x^{2}} \)
Use the product rule:
\[
w'(x) = 6e^{-x^{2}} + 6x \cdot (-2x)e^{-x^{2}} = 6e^{-x^{2}} -12x^{2}e^{-x^{2}} = 6e^{-x^{2}}(1 - 2x^{2})
\]
Now, compute the second derivative:
\[
\frac{d^{2}z}{dx^{2}} = w'(x)y(x) + w(x)y'(x)
\]
Compute \( y'(x) \):
\[
y'(x) = \frac{d}{dx}[1 - x^{2}] = -2x
\]
Substitute back:
\[
\frac{d^{2}z}{dx^{2}} = [6e^{-x^{2}}(1 - 2x^{2})](1 - x^{2}) + [6xe^{-x^{2}}](-2x)
\]
Simplify the second term:
\[
6xe^{-x^{2}}(-2x) = -12x^{2}e^{-x^{2}}
\]
Now, combine terms:
\[
\frac{d^{2}z}{dx^{2}} = 6e^{-x^{2}}(1 - 2x^{2})(1 - x^{2}) - 12x^{2}e^{-x^{2}}
\]
Factor out \( 6e^{-x^{2}} \):
\[
\frac{d^{2}z}{dx^{2}} = 6e^{-x^{2}}\left[(1 - 2x^{2})(1 - x^{2}) - 2x^{2}\right]
\]
Simplify the expression inside the brackets:
First, expand \( (1 - 2x^{2})(1 - x^{2}) \):
\[
(1 - 2x^{2})(1 - x^{2}) = (1)(1 - x^{2}) - 2x^{2}(1 - x^{2}) = (1 - x^{2}) - 2x^{2} + 2x^{4} = 1 - x^{2} - 2x^{2} + 2x^{4} = 1 - 3x^{2} + 2x^{4}
\]
Now, subtract \( 2x^{2} \):
\[
1 - 3x^{2} + 2x^{4} - 2x^{2} = 1 - 5x^{2} + 2x^{4}
\]
So, the second derivative is:
\[
\frac{d^{2}z}{dx^{2}} = 6e^{-x^{2}}(1 - 5x^{2} + 2x^{4})
\]
**Evaluate the second derivative at each critical point:**
1. **At \( x = 0 \):**
\[
\frac{d^{2}z}{dx^{2}} = 6e^{0}(1 - 0 + 0) = 6 \cdot 1 \cdot 1 = 6 > 0
\]
**Conclusion:** \( x = 0 \) is a **local minimum**.
2. **At \( x = 1 \):**
\[
\frac{d^{2}z}{dx^{2}} = 6e^{-1^{2}}(1 - 5(1)^{2} + 2(1)^{4}) = 6e^{-1}(1 - 5 + 2) = 6e^{-1}(-2) = -12e^{-1} < 0
\]
**Conclusion:** \( x = 1 \) is a **local maximum**.
3. **At \( x = -1 \):**
\[
\frac{d^{2}z}{dx^{2}} = 6e^{-(-1)^{2}}(1 - 5(-1)^{2} + 2(-1)^{4}) = 6e^{-1}(1 - 5 + 2) = 6e^{-1}(-2) = -12e^{-1} < 0
\]
**Conclusion:** \( x = -1 \) is a **local maximum**.
---
### Step 3: Analyze the Function's Behavior
**Function Values at Critical Points:**
- At \( x = 0 \):
\[
z(0) = 3(0)^{2}e^{0} = 0
\]
- At \( x = \pm 1 \):
\[
z(\pm 1) = 3(1)^{2}e^{-1^{2}} = 3e^{-1} \approx 1.1036
\]
**Behavior as \( x \to \pm \infty \):**
As \( x \to \pm \infty \):
- \( e^{-x^{2}} \to 0 \) because the exponent \( -x^{2} \) becomes large and negative.
- \( x^{2}e^{-x^{2}} \to 0 \) because \( x^{2} \) grows polynomially, but \( e^{-x^{2}} \) decays exponentially faster.
Therefore:
\[
\lim_{x \to \pm \infty} z(x) = 0
\]
**Graphical Interpretation:**
- The function \( z(x) \) is **always non-negative**, as both \( x^{2} \) and \( e^{-x^{2}} \) are non-negative for all real \( x \).
- It has a **local minimum** at \( x = 0 \) with \( z(0) = 0 \).
- It has **local maxima** at \( x = \pm 1 \) with \( z(\pm 1) = 3e^{-1} \).
- The function is **symmetric** about the \( x \)-axis (even function), since \( z(-x) = z(x) \).
- As \( x \) moves away from \( x = \pm 1 \), \( z(x) \) decreases towards zero.
---
### **Summary**
- **Number of Local Maxima:** **2** (at \( x = \pm 1 \))
- **Number of Local Minima:** **1** (at \( x = 0 \))
- The function approaches zero as \( x \to \pm \infty \).
- The function is **even** and **always non-negative**.
**Visualization Tip:** Plotting \( z(x) \) will show two peaks at \( x = \pm 1 \) and a valley at \( x = 0 \), with the function approaching zero as \( x \) moves towards positive or negative infinity. This confirms our analytical findings.
---
### **Conclusion**
By taking the derivative and using the second derivative test, we've determined that \( z(x) = 3x^{2}e^{-x^{2}} \) has:
- A local minimum at \( x = 0 \).
- Local maxima at \( x = \pm 1 \).
- The function is symmetric about the \( y \)-axis and approaches zero as \( x \to \pm \infty \).
This analysis provides a clear understanding of the function's shape and critical points.