Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`f(x)=sin^(4)x+cos^(4)x,0ltxlt(pi)/(2).`

Answer

To find the points of local maxima and minima of the function \( f(x) = \sin^4 x + \cos^4 x \) for \( 0 < x < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)

We start by differentiating the function:

\[ f'(x) = \frac{d}{dx}(\sin^4 x) + \frac{d}{dx}(\cos^4 x) \]

Using the chain rule, we have:

\[ \frac{d}{dx}(\sin^4 x) = 4 \sin^3 x \cdot \cos x \]
\[ \frac{d}{dx}(\cos^4 x) = 4 \cos^3 x \cdot (-\sin x) = -4 \cos^3 x \cdot \sin x \]

So,

\[ f'(x) = 4 \sin^3 x \cos x - 4 \cos^3 x \sin x \]

Factoring out \( 4 \sin x \cos x \):

\[ f'(x) = 4 \sin x \cos x (\sin^2 x - \cos^2 x) \]

### Step 2: Set the first derivative to zero

To find critical points, we set \( f'(x) = 0 \):

\[ 4 \sin x \cos x (\sin^2 x - \cos^2 x) = 0 \]

This gives us two cases to consider:

1. \( \sin x = 0 \)
2. \( \sin^2 x - \cos^2 x = 0 \)

### Step 3: Solve for critical points

1. **From \( \sin x = 0 \)**:
- In the interval \( 0 < x < \frac{\pi}{2} \), \( \sin x = 0 \) does not yield any valid solutions.

2. **From \( \sin^2 x - \cos^2 x = 0 \)**:
- This simplifies to \( \sin^2 x = \cos^2 x \) or \( \tan^2 x = 1 \).
- Thus, \( \tan x = 1 \) implies \( x = \frac{\pi}{4} \).

### Step 4: Determine the nature of the critical point

To determine whether \( x = \frac{\pi}{4} \) is a local maximum or minimum, we need to find the second derivative \( f''(x) \).

### Step 5: Find the second derivative \( f''(x) \)

Differentiating \( f'(x) \):

\[ f'(x) = 4 \sin x \cos x (\sin^2 x - \cos^2 x) \]

Using the product rule and simplifying, we find:

\[ f''(x) = 4 \left( \cos^2 x (\sin^2 x - \cos^2 x) + \sin^2 x \cdot 2\sin x \cos x - \sin^2 x \cdot 2\sin x \cos x \right) \]

Evaluating at \( x = \frac{\pi}{4} \):

\[ f''\left(\frac{\pi}{4}\right) = 4 \left( \cos^2\left(\frac{\pi}{4}\right) \cdot 0 + \sin^2\left(\frac{\pi}{4}\right) \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right) \right) \]

Since \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):

\[ f''\left(\frac{\pi}{4}\right) = 4 \left( 0 + \left(\frac{1}{\sqrt{2}}\right)^2 \cdot 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \right) = 4 \cdot \frac{1}{2} \cdot 2 \cdot \frac{1}{2} = 2 > 0 \]

Since \( f''\left(\frac{\pi}{4}\right) > 0 \), \( x = \frac{\pi}{4} \) is a local minimum.

### Step 6: Find the local minimum value

Now, we calculate \( f\left(\frac{\pi}{4}\right) \):

\[ f\left(\frac{\pi}{4}\right) = \sin^4\left(\frac{\pi}{4}\right) + \cos^4\left(\frac{\pi}{4}\right) = 2 \left(\frac{1}{\sqrt{2}}\right)^4 = 2 \cdot \frac{1}{4} = \frac{1}{2} \]

### Conclusion

The function \( f(x) = \sin^4 x + \cos^4 x \) has a local minimum at \( x = \frac{\pi}{4} \) with a minimum value of \( \frac{1}{2} \).