If `f(x)=cos^(-1)(4x^3-3x)and lim_(xto1/2+)f'(x)=a and lim_(xto1/2-)f'(x)=b` then `a + b+ 3` is equal to ____

Answer

To solve the problem, we need to find the limits of the derivative of the function \( f(x) = \cos^{-1}(4x^3 - 3x) \) as \( x \) approaches \( \frac{1}{2} \) from the right and left. We will denote these limits as \( a \) and \( b \) respectively, and then compute \( a + b + 3 \).

### Step 1: Rewrite the function using a trigonometric identity
We start with the function:
\[ f(x) = \cos^{-1}(4x^3 - 3x) \]
Notice that \( 4x^3 - 3x \) can be expressed in terms of cosine:
\[ 4x^3 - 3x = \cos(3\theta) \quad \text{where } \theta = \cos^{-1}(x) \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = \cos^{-1}(\cos(3\theta)) = 3\theta = 3\cos^{-1}(x) \]

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(3\cos^{-1}(x)) = 3 \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{3}{\sqrt{1 - x^2}} \]

### Step 3: Evaluate the limits as \( x \) approaches \( \frac{1}{2} \)
Now we need to find the limits \( a \) and \( b \):
\[ \lim_{x \to \frac{1}{2}^+} f'(x) = -\frac{3}{\sqrt{1 - \left(\frac{1}{2}\right)^2}} = -\frac{3}{\sqrt{1 - \frac{1}{4}}} = -\frac{3}{\sqrt{\frac{3}{4}}} = -\frac{3}{\frac{\sqrt{3}}{2}} = -\frac{6}{\sqrt{3}} = -2\sqrt{3} \]
Thus, \( a = -2\sqrt{3} \).

Now, we calculate the left-hand limit:
\[ \lim_{x \to \frac{1}{2}^-} f'(x) = -\frac{3}{\sqrt{1 - \left(\frac{1}{2}\right)^2}} = -\frac{3}{\sqrt{1 - \frac{1}{4}}} = -\frac{3}{\sqrt{\frac{3}{4}}} = -\frac{6}{\sqrt{3}} = -2\sqrt{3} \]
Thus, \( b = -2\sqrt{3} \).

### Step 4: Calculate \( a + b + 3 \)
Now we can compute \( a + b + 3 \):
\[ a + b + 3 = -2\sqrt{3} - 2\sqrt{3} + 3 = -4\sqrt{3} + 3 \]

### Final Answer
The final answer is:
\[ \boxed{3 - 4\sqrt{3}} \]