Find the realtionship between a and b so that the function defined by `f(x)={{:(,ax+1,x le 3),(,bx+3, x gt 3):}" is continuous at x=3"`

Answer

To find the relationship between \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} ax + 1 & \text{if } x \leq 3 \\ bx + 3 & \text{if } x > 3 \end{cases} \]

is continuous at \( x = 3 \), we will use the definition of continuity. A function is continuous at a point if the following condition holds:

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

where \( c \) is the point of interest, in this case, \( c = 3 \).

### Step 1: Calculate \( \lim_{x \to 3^-} f(x) \)

For \( x \) approaching 3 from the left (i.e., \( x \leq 3 \)), we use the first part of the piecewise function:

\[ f(x) = ax + 1 \]

Thus,

\[ \lim_{x \to 3^-} f(x) = a(3) + 1 = 3a + 1 \]

### Step 2: Calculate \( \lim_{x \to 3^+} f(x) \)

For \( x \) approaching 3 from the right (i.e., \( x > 3 \)), we use the second part of the piecewise function:

\[ f(x) = bx + 3 \]

Thus,

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

### Step 3: Calculate \( f(3) \)

Since \( f(3) \) is defined by the first case (as \( 3 \leq 3 \)):

\[ f(3) = a(3) + 1 = 3a + 1 \]

### Step 4: Set the limits equal for continuity

For the function to be continuous at \( x = 3 \), we need:

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

This gives us the equations:

\[ 3a + 1 = 3b + 3 \]

### Step 5: Solve for the relationship between \( a \) and \( b \)

Rearranging the equation:

\[ 3a + 1 = 3b + 3 \]

Subtract \( 3b \) and \( 1 \) from both sides:

\[ 3a - 3b = 3 - 1 \]

Simplifying gives:

\[ 3a - 3b = 2 \]

Dividing the entire equation by 3:

\[ a - b = \frac{2}{3} \]

Thus, the relationship between \( a \) and \( b \) is:

\[ a = b + \frac{2}{3} \]