If `f(x) ={(lambdasqrt(2x+3),0 lex le3),(mux + 12, 3 < x le 9):}` is differentiable at `x = 3`, then the value of `lambda + mu` is equal to

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable at \( x = 3 \). The function is defined as follows:

\[ f(x) = \begin{cases} \lambda \sqrt{2x + 3} & \text{for } 0 \leq x \leq 3 \\ \mu x + 12 & \text{for } 3 < x \leq 9 \end{cases} \]

### Step 1: Ensure Continuity at \( x = 3 \)

For \( f(x) \) to be continuous at \( x = 3 \), the left-hand limit as \( x \) approaches 3 must equal the right-hand limit at \( x = 3 \):

\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to 3^-} f(x) = \lambda \sqrt{2(3) + 3} = \lambda \sqrt{9} = 3\lambda \]

Calculating the right-hand limit:

\[ \lim_{x \to 3^+} f(x) = \mu(3) + 12 = 3\mu + 12 \]

Setting these two limits equal gives us the first equation:

\[ 3\lambda = 3\mu + 12 \quad \text{(1)} \]

### Step 2: Ensure Differentiability at \( x = 3 \)

For \( f(x) \) to be differentiable at \( x = 3 \), the left-hand derivative must equal the right-hand derivative:

\[ \lim_{x \to 3^-} f'(x) = \lim_{x \to 3^+} f'(x) \]

Calculating the left-hand derivative:

\[ f'(x) = \frac{\lambda}{2\sqrt{2x + 3}} \cdot 2 = \frac{\lambda}{\sqrt{2x + 3}} \quad \text{for } 0 \leq x < 3 \]
At \( x = 3 \):

\[ \lim_{x \to 3^-} f'(x) = \frac{\lambda}{\sqrt{2(3) + 3}} = \frac{\lambda}{\sqrt{9}} = \frac{\lambda}{3} \]

Calculating the right-hand derivative:

\[ f'(x) = \mu \quad \text{for } 3 < x \leq 9 \]
Thus:

\[ \lim_{x \to 3^+} f'(x) = \mu \]

Setting these two derivatives equal gives us the second equation:

\[ \frac{\lambda}{3} = \mu \quad \text{(2)} \]

### Step 3: Solve the System of Equations

From equation (2), we can express \( \mu \) in terms of \( \lambda \):

\[ \mu = \frac{\lambda}{3} \]

Substituting this into equation (1):

\[ 3\lambda = 3\left(\frac{\lambda}{3}\right) + 12 \]

This simplifies to:

\[ 3\lambda = \lambda + 12 \]

Rearranging gives:

\[ 3\lambda - \lambda = 12 \implies 2\lambda = 12 \implies \lambda = 6 \]

Now substituting \( \lambda = 6 \) back into equation (2) to find \( \mu \):

\[ \mu = \frac{6}{3} = 2 \]

### Step 4: Find \( \lambda + \mu \)

Now we can find \( \lambda + \mu \):

\[ \lambda + \mu = 6 + 2 = 8 \]

### Final Answer

Thus, the value of \( \lambda + \mu \) is \( \boxed{8} \).