If ` f(x) = sqrt(x^2 - 4x +4)` on the interval [0.3] then ` f' (2 +) = . . .. . And f '(2-) =…..`

Answer

To find the right-hand derivative \( f'(2^+) \) and the left-hand derivative \( f'(2^-) \) of the function \( f(x) = \sqrt{x^2 - 4x + 4} \), we will follow these steps:

### Step 1: Simplify the Function
First, we simplify the function \( f(x) \).

\[ f(x) = \sqrt{x^2 - 4x + 4} \]
We can factor the expression inside the square root:
\[ x^2 - 4x + 4 = (x - 2)^2 \]
Thus, we have:
\[ f(x) = \sqrt{(x - 2)^2} \]
This simplifies to:
\[ f(x) = |x - 2| \]

### Step 2: Determine the Right-Hand Derivative \( f'(2^+) \)
The right-hand derivative at \( x = 2 \) is calculated using the limit definition:
\[ f'(2^+) = \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} \]
Substituting \( f(x) \):
\[ f(2 + h) = |(2 + h) - 2| = |h| = h \quad (\text{since } h > 0) \]
And we know:
\[ f(2) = |2 - 2| = 0 \]
Now substituting into the limit:
\[ f'(2^+) = \lim_{h \to 0^+} \frac{h - 0}{h} = \lim_{h \to 0^+} 1 = 1 \]

### Step 3: Determine the Left-Hand Derivative \( f'(2^-) \)
The left-hand derivative at \( x = 2 \) is calculated similarly:
\[ f'(2^-) = \lim_{h \to 0^+} \frac{f(2 - h) - f(2)}{-h} \]
Substituting \( f(x) \):
\[ f(2 - h) = |(2 - h) - 2| = |-h| = h \quad (\text{since } h > 0) \]
Again, we know:
\[ f(2) = 0 \]
Now substituting into the limit:
\[ f'(2^-) = \lim_{h \to 0^+} \frac{h - 0}{-h} = \lim_{h \to 0^+} -1 = -1 \]

### Final Answers
Thus, we have:
\[ f'(2^+) = 1 \quad \text{and} \quad f'(2^-) = -1 \]