Questions: The function f(x)= 4-x^2 x leq 1 m x+b x>1 is continuous and differentiable for all real numbers. What must be the values of m and b?

Transcript text: The function \[ f(x)=\left\{\begin{array}{ll} 4-x^{2} & x \leq 1 \\ m x+b & x>1 \end{array}\right. \] is continuous and differentiable for all real numbers. What must be the values of $m$ and $b$ ?

Solution

Solution Steps

To ensure the function is continuous and differentiable at $ x = 1 $, we need to match the value and the derivative of the two pieces of the function at $ x = 1 $.

Continuity at $ x = 1 $: The left-hand limit as $ x $ approaches 1 from the left must equal the right-hand limit as $ x $ approaches 1 from the right.
Differentiability at $ x = 1 $: The derivative from the left must equal the derivative from the right at $ x = 1 $.

Step 1: Continuity Condition

To ensure continuity at $ x = 1 $, we set the left-hand limit equal to the right-hand limit: \[ 4 - 1^2 = m \cdot 1 + b \] This simplifies to: \[ 3 = m + b \]

Step 2: Differentiability Condition

To ensure differentiability at $ x = 1 $, we set the derivative from the left equal to the derivative from the right: \[ \frac{d}{dx}(4 - x^2) \bigg|_{x=1} = \frac{d}{dx}(m \cdot x + b) \bigg|_{x=1} \] Calculating the derivatives gives: \[ -2 = m \]

Step 3: Solve for $ b $

Substituting $ m = -2 $ into the continuity equation: \[ 3 = -2 + b \] Solving for $ b $ yields: \[ b = 5 \]