Let `f(x) = 1 + 4x - x^(2), AA x in R` `g(x) = max {f(t), x le t le (x + 1), 0 le x lt 3min {(x + 3), 3 le x le 5}` Verify conntinuity of g(x), for all `x in [0, 5]`

Answer

To verify the continuity of the function \( g(x) \) defined as

\[ g(x) = \max \{ f(t) \,|\, x \leq t \leq x+1 \} \quad \text{for } 0 \leq x < 3 \]
and
\[ g(x) = \min \{ x + 3 \,|\, 3 \leq x \leq 5 \} \]

where \( f(x) = 1 + 4x - x^2 \), we will follow these steps:

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) = 1 + 4x - x^2 \) is a quadratic function. To find its maximum value, we can differentiate it.

\[ f'(x) = 4 - 2x \]

Setting the derivative to zero to find critical points:

\[ 4 - 2x = 0 \implies x = 2 \]

### Step 2: Determine the maximum value of \( f(t) \) in the interval \( [x, x+1] \)

- For \( x < 2 \):
- The maximum occurs at \( t = x + 1 \).

\[ g(x) = f(x + 1) = 1 + 4(x + 1) - (x + 1)^2 \]

Calculating \( g(x) \):

\[ g(x) = 1 + 4x + 4 - (x^2 + 2x + 1) = 4 + 2x - x^2 \]

- For \( x = 2 \):

\[ g(2) = f(2) = 1 + 4(2) - 2^2 = 1 + 8 - 4 = 5 \]

- For \( x > 2 \) and \( x < 3 \):
- The maximum occurs at \( t = 2 \) since \( t \) will be in the interval \( [x, x+1] \).

\[ g(x) = f(2) = 5 \]

### Step 3: Define \( g(x) \) for \( 3 \leq x \leq 5 \)

For \( 3 \leq x \leq 5 \):

\[ g(x) = x + 3 \]

### Step 4: Check continuity at the transition points \( x = 2 \) and \( x = 3 \)

- At \( x = 2 \):

\[ g(2) = 5 \quad \text{(from the first case)} \]

- As \( x \) approaches 2 from the left (\( x \to 2^- \)):

\[ g(2^-) = 5 \]

- As \( x \) approaches 2 from the right (\( x \to 2^+ \)):

\[ g(2^+) = 5 \]

Thus, \( g(x) \) is continuous at \( x = 2 \).

- At \( x = 3 \):

\[ g(3) = 3 + 3 = 6 \quad \text{(from the second case)} \]

- As \( x \) approaches 3 from the left (\( x \to 3^- \)):

\[ g(3^-) = 5 \]

- As \( x \) approaches 3 from the right (\( x \to 3^+ \)):

\[ g(3^+) = 6 \]

Thus, \( g(x) \) is discontinuous at \( x = 3 \).

### Conclusion

The function \( g(x) \) is continuous on the intervals \( [0, 2] \) and \( [3, 5] \) but is discontinuous at \( x = 3 \).