The value of b for which the function
`f(x) = {{:(5x-4, 0 lt x le 1),(4x^(2) + 3bx, 1 lt x lt 2):}` is continuous at every point of its domain, is

Options

A. `-1`
B. 0
C. 1
D. `13//3`

Answer

To find the value of \( b \) for which the function

\[ f(x) = \begin{cases} 5x - 4 & \text{for } 0 < x \leq 1 \\ 4x^2 + 3bx & \text{for } 1 < x < 2 \end{cases} \]

is continuous at every point of its domain, we need to ensure that the function is continuous at the point where the two pieces meet, which is at \( x = 1 \).

### Step-by-step Solution:

1. **Evaluate \( f(1) \)**:
Since \( 1 \) falls in the first piece of the function, we use:
\[ f(1) = 5(1) - 4 = 5 - 4 = 1 \]

2. **Find the limit as \( x \) approaches \( 1 \) from the right**:
For \( x \) approaching \( 1 \) from the right, we use the second piece of the function:
\[ \lim_{x \to 1^+} f(x) = 4(1)^2 + 3b(1) = 4 + 3b \]

3. **Set the limit equal to \( f(1) \)**:
For continuity at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]
Thus, we set:
\[ 4 + 3b = 1 \]

4. **Solve for \( b \)**:
Rearranging the equation gives:
\[ 3b = 1 - 4 \]
\[ 3b = -3 \]
\[ b = -1 \]

### Conclusion:
The value of \( b \) for which the function is continuous at every point of its domain is \( b = -1 \).