Math Problem Statement

Let f be defined by the function f(x)={3-x, x<1. ax^2 + bx, x≥1. A.)If the function is continuous at. X=1, what is the relationship between an and B explain your reasoning using limits. B.) find the unique values of an and bthat will make both continuous and differentiable at x=1. Show your analysis using limits.

Solution

A) Continuity at x=1x = 1

For the function f(x)f(x) to be continuous at x=1x = 1, the left-hand limit (LHL) and right-hand limit (RHL) at x=1x = 1 must be equal to the value of the function at x=1x = 1.

1. Left-hand limit (LHL) as x1x \to 1^-:

The piece of the function when x<1x < 1 is given by: f(x)=3xf(x) = 3 - x

As x1x \to 1^-: limx1f(x)=31=2\lim_{x \to 1^-} f(x) = 3 - 1 = 2

2. Right-hand limit (RHL) as x1+x \to 1^+:

The piece of the function when x1x \geq 1 is given by: f(x)=ax2+bxf(x) = ax^2 + bx

As x1+x \to 1^+: limx1+f(x)=a(1)2+b(1)=a+b\lim_{x \to 1^+} f(x) = a(1)^2 + b(1) = a + b

For continuity, the limits must match: limx1f(x)=limx1+f(x)\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) 2=a+b2 = a + b

Thus, the relationship between aa and bb is: a+b=2a + b = 2


B) Continuity and Differentiability at x=1x = 1

For f(x)f(x) to be differentiable at x=1x = 1, the derivatives from the left and right must also be equal.

1. Left-hand derivative (LHD) as x1x \to 1^-:

The derivative of f(x)=3xf(x) = 3 - x when x<1x < 1 is: f(x)=1f'(x) = -1

Thus: limx1f(x)=1\lim_{x \to 1^-} f'(x) = -1

2. Right-hand derivative (RHD) as x1+x \to 1^+:

The derivative of f(x)=ax2+bxf(x) = ax^2 + bx when x1x \geq 1 is: f(x)=2ax+bf'(x) = 2ax + b

As x1+x \to 1^+: limx1+f(x)=2a(1)+b=2a+b\lim_{x \to 1^+} f'(x) = 2a(1) + b = 2a + b

For differentiability, the left and right derivatives must match: limx1f(x)=limx1+f(x)\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) 1=2a+b-1 = 2a + b

3. System of Equations:

From continuity: a+b=2a + b = 2

From differentiability: 2a+b=12a + b = -1

Subtract the first equation from the second: (2a+b)(a+b)=12(2a + b) - (a + b) = -1 - 2 a=3a = -3

Substitute a=3a = -3 into a+b=2a + b = 2: 3+b=2-3 + b = 2 b=5b = 5


Final Answer:

The unique values of aa and bb that make f(x)f(x) both continuous and differentiable at x=1x = 1 are: a=3,b=5a = -3, \quad b = 5

Verification:

  1. Continuity: a+b=3+5=2(satisfied)a + b = -3 + 5 = 2 \quad \text{(satisfied)}

  2. Differentiability: 2a+b=2(3)+5=6+5=1(satisfied)2a + b = 2(-3) + 5 = -6 + 5 = -1 \quad \text{(satisfied)}

Would you like further clarification on any step or additional examples?


Follow-Up Questions:

  1. Why does continuity imply matching function values at x=1x = 1?
  2. How does differentiability rely on the function's derivative?
  3. What are the implications of discontinuity at x=1x = 1?
  4. How does this process change if f(x)f(x) were a piecewise-defined trigonometric function?
  5. What role does the smoothness of polynomials play in differentiability?

Tip:

Always check both continuity and differentiability when working with piecewise functions at critical points.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Continuity
Differentiability
Limits

Formulas

Limit definition for continuity: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \)
Limit definition for differentiability: \( \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) \)

Theorems

Definition of Continuity
Definition of Differentiability

Suitable Grade Level

Grades 11-12 or early college calculus