Question

Consider the given function \( f(x) \). \[ f(x) = \begin{cases} ax + b & \text{for } x < 1 \\ x^3 + x^2 + 1 & \text{for } x \geq 1 \end{cases} \] If the function is differentiable everywhere, the value of \( b \) must be __________ (rounded off to one decimal place).

Hint:

For piecewise functions to be differentiable everywhere, they must be continuous and have equal derivatives at the point where the function changes its definition. Always check both the continuity and differentiability conditions at that point.

Answer 1

Correct Answer: -2.1

For the function \( f(x) \) to be differentiable everywhere, it must be both continuous and differentiable at \( x = 1 \) (the point where the piecewise function changes). Step 1: Continuity Condition
First, let's ensure that the function is continuous at \( x = 1 \). For continuity at \( x = 1 \), we need the left-hand limit and the right-hand limit to be equal at \( x = 1 \). The function for \( x<1 \) is \( f(x) = ax + b \).
The function for \( x \geq 1 \) is \( f(x) = x^3 + x^2 + 1 \).
We want the limits from both sides to match at \( x = 1 \):
Left-hand limit as \( x \to 1^- \): \[ \lim_{x \to 1^-} f(x) = a(1) + b = a + b \] Right-hand limit as \( x \to 1^+ \): \[ \lim_{x \to 1^+} f(x) = 1^3 + 1^2 + 1 = 1 + 1 + 1 = 3 \] For continuity at \( x = 1 \), we require: \[ a + b = 3 \] So, we have the equation: \[ a + b = 3 \quad {(Equation 1)} \] Step 2: Differentiability Condition
Next, we ensure that the function is differentiable at \( x = 1 \). For differentiability at \( x = 1 \), the derivative from the left must be equal to the derivative from the right. The derivative of \( f(x) = ax + b \) for \( x<1 \) is: \[ f'(x) = a \] The derivative of \( f(x) = x^3 + x^2 + 1 \) for \( x \geq 1 \) is: \[ f'(x) = 3x^2 + 2x \] At \( x = 1 \), we want the left-hand derivative and the right-hand derivative to be equal: \[ a = 3(1)^2 + 2(1) = 3 + 2 = 5 \] So, from the differentiability condition, we have: \[ a = 5 \quad {(Equation 2)} \] Step 3: Solving for \( b \)\ From Equation 2, we know \( a = 5 \). Substituting \( a = 5 \) into Equation 1: \[ 5 + b = 3 \] Solving for \( b \): \[ b = 3 - 5 = -2 \] Thus, the value of \( b \) is -2.