To find $f'(1)$, where $f(x) = \sqrt{4x - 5}$ and $x \geq -1.25$, we need to compute the derivative of $f(x)$ and then evaluate it at $x = 1$.

Step 1: Differentiate $f(x) = \sqrt{4x - 5}$

We can rewrite the function as: $f(x) = (4x - 5)^{1/2}$

Now, apply the chain rule to differentiate: $f'(x) = \frac{d}{dx} \left( (4x - 5)^{1/2} \right)$

The derivative of $(4x - 5)^{1/2}$ is: $f'(x) = \frac{1}{2}(4x - 5)^{-1/2} \cdot \frac{d}{dx}(4x - 5)$

The derivative of $4x - 5$ is simply 4, so: $f'(x) = \frac{1}{2}(4x - 5)^{-1/2} \cdot 4$

Simplifying: $f'(x) = \frac{2}{\sqrt{4x - 5}}$

Step 2: Evaluate $f'(x)$ at $x = 1$

Now, substitute $x = 1$ into the expression for $f'(x)$: $f'(1) = \frac{2}{\sqrt{4(1) - 5}} = \frac{2}{\sqrt{4 - 5}} = \frac{2}{\sqrt{-1}}$

Since $\sqrt{-1}$ is imaginary, this result suggests that $f'(1)$ is undefined in the real number system. The derivative at $x = 1$ does not exist because the expression inside the square root becomes negative, indicating a complex value.

Conclusion:

$f'(1)$ does not exist in the real number domain because $\sqrt{-1}$ is not defined in the real numbers.

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Would you like more details or clarification?

Here are some related questions you might find interesting:

1. How do you differentiate square root functions in general?
2. How do we handle derivatives involving complex numbers?
3. What is the domain of a function like $f(x) = \sqrt{4x - 5}$?
4. What happens to the derivative if the argument inside the square root becomes negative?
5. How can we calculate the derivative for functions involving square roots of expressions other than linear ones?

Tip: Always check the domain of a function before performing derivative calculations. If the domain leads to undefined operations (like taking the square root of a negative number), the derivative may not exist at those points.