0
To determine the derivative \(\frac{dy}{dx}\) of the function \(f(x) = |x-3| + |x-4|\) at \(x = 3.5\), let's proceed with the following step-by-step approach:
Understanding the Function
The function is given as:
\(f(x) = |x-3| + |x-4|\)
This function involves absolute values, which can be expressed as piecewise functions. Let's break it into different intervals:
- Interval \(x < 3\): \(|x-3| = -(x-3)\) and \(|x-4| = -(x-4)\)
- Interval \(3 \leq x < 4\): \(|x-3| = (x-3)\) and \(|x-4| = -(x-4)\)
- Interval \(x \geq 4\): \(|x-3| = (x-3)\) and \(|x-4| = (x-4)\)
Calculating Derivative at \(x = 3.5\)
Since \(3 \leq x < 4\) when \(x = 3.5\), we consider this interval:
\(|x-3| = (x-3)\) and \(|x-4| = -(x-4)\)
Hence, on this interval:
\(f(x) = (x-3) - (x-4) = x - 3 - x + 4 = 1\)
The derivative of a constant function, such as \(1\), is:
\(\frac{d}{dx}(1) = 0\)
Conclusion
Thus, the value of \(\frac{dy}{dx}\) at \(x = 3.5\) is 0.
This result is obtained because at \(x = 3.5\), the function \(f(x)\) is a constant value, and any changes to \(x\) around this point do not result in a change in \(f(x)\). Therefore, the correct answer is \( \boxed{0} \).
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Select the correct answer using the code given below:
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