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.

### A) Continuity at \(x = 1\)

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

#### 1. **Left-hand limit (LHL) as \(x \to 1^-\):**

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

As \(x \to 1^-\):
\[
\lim_{x \to 1^-} f(x) = 3 - 1 = 2
\]

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

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

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

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

Thus, the relationship between \(a\) and \(b\) is:
\[
a + b = 2
\]

---

### B) Continuity and Differentiability at \(x = 1\)

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

#### 1. **Left-hand derivative (LHD) as \(x \to 1^-\):**

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

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

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

The derivative of \(f(x) = ax^2 + bx\) when \(x \geq 1\) is:
\[
f'(x) = 2ax + b
\]

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

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

#### 3. **System of Equations:**

From continuity:
\[
a + b = 2
\]

From differentiability:
\[
2a + b = -1
\]

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

Substitute \(a = -3\) into \(a + b = 2\):
\[
-3 + b = 2
\]
\[
b = 5
\]

---

### Final Answer:
The unique values of \(a\) and \(b\) that make \(f(x)\) both continuous and differentiable at \(x = 1\) are:
\[
a = -3, \quad b = 5
\]

### Verification:

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

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

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

---

### Follow-Up Questions:
1. Why does continuity imply matching function values at \(x = 1\)?
2. How does differentiability rely on the function's derivative?
3. What are the implications of discontinuity at \(x = 1\)?
4. How does this process change if \(f(x)\) 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.

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 a and b? Explain your reasoning using limits. B.) Find the unique values of a and b that will make both continuous and differentiable at x=1. Show your analysis using limits.

Explore how to determine the values of constants a and b to ensure the piecewise function f(x) is continuous and differentiable at x=1. This solution uses limit definitions to verify continuity and differentiability, making it suitable for students in Grades 11-12 or early college calculus.

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