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We are given a piecewise function \( f(x) \):

\[
f(x) = 
\begin{cases} 
x^2 - 2x - a & \text{if } x < 1, \\
-x^2 + bx + 2 & \text{if } x \geq 1,
\end{cases}
\]
and we need to find the values of \( a \) and \( b \) such that \( f(x) \) is continuous and differentiable at \( x = 1 \).

### Step 1: Ensure Continuity at \( x = 1 \)
For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit and right-hand limit must be equal to \( f(1) \). This means:

\[
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1).
\]

1. From the first piece of the function, when \( x < 1 \):
   \[
   \lim_{x \to 1^-} f(x) = 1^2 - 2(1) - a = 1 - 2 - a = -1 - a.
   \]
   
2. From the second piece of the function, when \( x \geq 1 \):
   \[
   f(1) = -(1)^2 + b(1) + 2 = -1 + b + 2 = b + 1.
   \]

To ensure continuity, we set the limits equal to each other:

\[
-1 - a = b + 1.
\]

This simplifies to:

\[
a + b = -2. \quad \text{(Equation 1)}
\]

### Step 2: Ensure Differentiability at \( x = 1 \)
For \( f(x) \) to be differentiable at \( x = 1 \), the derivatives from both sides must be equal at \( x = 1 \). So we first find the derivatives of the two pieces of the function.

1. For \( x < 1 \), \( f(x) = x^2 - 2x - a \). The derivative is:
   \[
   f'(x) = 2x - 2.
   \]
   Evaluating at \( x = 1 \):
   \[
   f'(1^-) = 2(1) - 2 = 0.
   \]

2. For \( x \geq 1 \), \( f(x) = -x^2 + bx + 2 \). The derivative is:
   \[
   f'(x) = -2x + b.
   \]
   Evaluating at \( x = 1 \):
   \[
   f'(1^+) = -2(1) + b = -2 + b.
   \]

To ensure differentiability, we set the derivatives equal at \( x = 1 \):

\[
0 = -2 + b.
\]

This simplifies to:

\[
b = 2. \quad \text{(Equation 2)}
\]

### Step 3: Solve for \( a \)
Now substitute \( b = 2 \) into Equation 1:

\[
a + 2 = -2,
\]

which gives:

\[
a = -4.
\]

### Final Answer
The values of \( a \) and \( b \) that make \( f(x) \) continuous and differentiable at \( x = 1 \) are:
\[
a = -4, \quad b = 2.
\]

Would you like a more detailed breakdown of any steps?

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Here are 5 related questions for further exploration:
1. What are the necessary conditions for a function to be continuous at a point?
2. How can you check the differentiability of a function at a point with a more complex piecewise function?
3. What happens if a function is continuous but not differentiable at a point?
4. How can the chain rule be used when differentiating piecewise functions with more complex boundaries?
5. How do second-order derivatives behave in piecewise functions at boundary points?

**Tip:** When dealing with piecewise functions, always ensure to check both continuity and differentiability by comparing limits and derivatives from both sides of the boundary.

Find the values of a and b that make f(x) continuous and differentiable at x = 1. The function is given as a piecewise function with f(x) = x^2 - 2x - a for x < 1, and f(x) = -x^2 + bx + 2 for x >= 1.

This problem focuses on determining the values of constants a and b that ensure a piecewise function is both continuous and differentiable at x = 1. The function contains quadratic expressions on both sides of the piecewise boundary, requiring the application of continuity and differentiability conditions.

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