Find the range of `f(x)=cos^4x+sin^2x-1`.

Explanation

To find the range of the function \( f(x) = \cos^4 x + \sin^2 x - 1 \), we can follow these steps:

### Step 1: Rewrite the function
We start with the given function:
\[ f(x) = \cos^4 x + \sin^2 x - 1 \]
We know that \( \sin^2 x = 1 - \cos^2 x \). Substituting this into the function gives:
\[ f(x) = \cos^4 x + (1 - \cos^2 x) - 1 \]
This simplifies to:
\[ f(x) = \cos^4 x - \cos^2 x \]

### Step 2: Let \( y = \cos^2 x \)
Now, let \( y = \cos^2 x \). Since \( \cos^2 x \) ranges from 0 to 1, we have:
\[ f(x) = y^2 - y \]
where \( 0 \leq y \leq 1 \).

### Step 3: Analyze the function \( g(y) = y^2 - y \)
Next, we analyze the quadratic function \( g(y) = y^2 - y \). This is a downward-opening parabola (since the coefficient of \( y^2 \) is positive).

### Step 4: Find the vertex of the parabola
To find the vertex of the parabola, we use the formula for the vertex \( y = -\frac{b}{2a} \):
\[ y = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]
Now, we evaluate \( g(y) \) at \( y = \frac{1}{2} \):
\[ g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]

### Step 5: Evaluate the endpoints
Next, we evaluate \( g(y) \) at the endpoints of the interval:
- At \( y = 0 \):
\[ g(0) = 0^2 - 0 = 0 \]
- At \( y = 1 \):
\[ g(1) = 1^2 - 1 = 0 \]

### Step 6: Determine the range
From our calculations:
- The minimum value of \( g(y) \) occurs at \( y = \frac{1}{2} \) and is \( -\frac{1}{4} \).
- The maximum values occur at the endpoints \( y = 0 \) and \( y = 1 \), both yielding \( 0 \).

Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{1}{4}, 0\right] \]

### Final Answer
The range of the function \( f(x) = \cos^4 x + \sin^2 x - 1 \) is:
\[ \boxed{[-\frac{1}{4}, 0]} \]