Let's analyze the continuity of the piecewise function f(x) defined as:

A function is continuous at a point if the left-hand limit, right-hand limit, and the function's value at that point all exist and are equal. For a piecewise function, we must check the points where the definition changes: x = 1, x = 3, and x = 5.

Step 1: Continuity at x = 1

Left-hand limit (x→1⁻): f(x) = 5 ⇒ limit = 5

Right-hand limit (x→1⁺): f(x) = a + bx ⇒ limit = a + b(1) = a + b

Function value at x=1: f(1) = 5

For continuity: a + b = 5 ...(Equation 1)

Step 2: Continuity at x = 3

Left-hand limit (x→3⁻): f(x) = a + bx ⇒ limit = a + b(3) = a + 3b

Right-hand limit (x→3⁺): f(x) = b + 5x ⇒ limit = b + 5(3) = b + 15

Function value at x=3: f(3) = b + 5(3) = b + 15

For continuity: a + 3b = b + 15 ⇒ a + 2b = 15 ...(Equation 2)

Step 3: Continuity at x = 5

Left-hand limit (x→5⁻): f(x) = b + 5x ⇒ limit = b + 5(5) = b + 25

Right-hand limit (x→5⁺): f(x) = 30 ⇒ limit = 30

Function value at x=5: f(5) = 30

For continuity: b + 25 = 30 ⇒ b = 5 ...(Equation 3)

Step 4: Solve the System of Equations

From Equation 3: b = 5

Substitute b = 5 into Equation 1: a + 5 = 5 ⇒ a = 0

Substitute b = 5 into Equation 2: a + 2(5) = 15 ⇒ a + 10 = 15 ⇒ a = 5

We get a contradiction: a must be both 0 (from Eq1) and 5 (from Eq2). Therefore, no values of a and b can satisfy all three continuity conditions simultaneously.

Final Answer

The function is not continuous for any values of a and b.

Related Topics

Continuity of Functions: A function f(x) is continuous at a point x = c if: $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)=f\left(c\right)$ All three values must exist and be equal.

Piecewise Functions: Functions defined by different expressions over different intervals. Special attention must be paid to the points where the definition changes to check for continuity.

Key Formulae

The fundamental condition for continuity at a point x = c: $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$