$$ x^4-4x^2+10 $$ find X- cordinates of all local minima using second derivative test

To find the local minima of the function 
\[
f(x) = x^4 - 4x^2 + 10,
\]
we will follow these steps using the second derivative test:

1. **Find the first derivative $ f'(x) $ and set it to zero to find critical points.**

2. **Find the second derivative $ f''(x) $ and evaluate it at the critical points.**

3. **Use the second derivative test to determine if the critical points are local minima.**

### Step 1: Find the first derivative
The first derivative of $ f(x) $ is:
\[
f'(x) = 4x^3 - 8x.
\]
Set $ f'(x) = 0 $ to find the critical points:
\[
4x^3 - 8x = 0 \implies 4x(x^2 - 2) = 0.
\]
Thus, the critical points are:
\[
x = 0, \quad x = \sqrt{2}, \quad x = -\sqrt{2}.
\]

### Step 2: Find the second derivative
The second derivative of $ f(x) $ is:
\[
f''(x) = 12x^2 - 8.
\]

### Step 3: Apply the second derivative test
Evaluate $ f''(x) $ at each critical point:
1. At $ x = 0 $:
   \[
   f''(0) = 12(0)^2 - 8 = -8.
   \]
   Since $ f''(0) < 0 $, this indicates a local maximum at $ x = 0 $.

2. At $ x = \sqrt{2} $:
   \[
   f''(\sqrt{2}) = 12(\sqrt{2})^2 - 8 = 12 \cdot 2 - 8 = 24 - 8 = 16.
   \]
   Since $ f''(\sqrt{2}) > 0 $, this indicates a local minimum at $ x = \sqrt{2} $.

3. At $ x = -\sqrt{2} $:
   \[
   f''(-\sqrt{2}) = 12(-\sqrt{2})^2 - 8 = 12 \cdot 2 - 8 = 16.
   \]
   Since $ f''(-\sqrt{2}) > 0 $, this indicates a local minimum at $ x = -\sqrt{2} $.

### Conclusion
The local minima of the function occur at the $ x $-coordinates:
\[
x = \sqrt{2} \text{ and } x = -\sqrt{2}.
\]

Would you like a detailed explanation of the second derivative test or have any further questions?

### Related Questions
1. How do you find local maxima and minima using the first derivative test?
2. What is the role of higher-order derivatives in determining the nature of critical points?
3. How does the shape of the graph near the critical points relate to the second derivative?
4. Can you have a local minimum or maximum where the second derivative is zero?
5. What are saddle points, and how do they differ from local minima and maxima?

### Tip
Always verify the nature of the critical points by considering the sign of the second derivative. If the second derivative changes signs around the critical point, it might indicate a point of inflection rather than a local extremum.

x^4 - 4x^2 + 10: Find X-coordinates of all local minima using the second derivative test.

Discover how to find the local minima of the function x^4 - 4x^2 + 10 using the second derivative test. This step-by-step solution covers first and second derivatives, critical points, and applying the second derivative test. Suitable for students in Grades 11-12 or college calculus courses.

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