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The problem presented asks to find values for \( a \) and \( b \) such that the given piecewise function is both continuous and differentiable at \( x = -2 \). The function is defined as:

\[
f(x) = 
\begin{cases} 
7x^3 - 2x^2 + 6, & \text{if } x \leq -2 \\
ax + b, & \text{if } x \geq -2
\end{cases}
\]

### Step 1: Ensuring Continuity at \( x = -2 \)
To ensure continuity at \( x = -2 \), we need the function values from both pieces to match at \( x = -2 \). That is:

\[
\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x)
\]

For \( x \leq -2 \), the value of \( f(x) \) at \( x = -2 \) is given by substituting into the first equation:

\[
f(-2) = 7(-2)^3 - 2(-2)^2 + 6 = -56 + 8 + 6 = -42
\]

For \( x \geq -2 \), the value of \( f(x) \) is \( ax + b \), so at \( x = -2 \):

\[
f(-2) = a(-2) + b = -2a + b
\]

To ensure continuity:

\[
-42 = -2a + b
\]
This is the first equation.

### Step 2: Ensuring Differentiability at \( x = -2 \)
To ensure differentiability at \( x = -2 \), the derivatives from both sides must match at \( x = -2 \). First, find the derivatives of each piece:

- For \( x \leq -2 \), \( f(x) = 7x^3 - 2x^2 + 6 \), so:

\[
f'(x) = 21x^2 - 4x
\]
At \( x = -2 \):

\[
f'(-2) = 21(-2)^2 - 4(-2) = 84 + 8 = 92
\]

- For \( x \geq -2 \), \( f(x) = ax + b \), so:

\[
f'(x) = a
\]

To ensure differentiability:

\[
92 = a
\]

### Step 3: Solving for \( b \)
Now that we have \( a = 92 \), substitute it into the continuity equation \( -42 = -2a + b \):

\[
-42 = -2(92) + b
\]
\[
-42 = -184 + b
\]
\[
b = -42 + 184 = 142
\]

### Final Answer:
The values of \( a \) and \( b \) that make the function both continuous and differentiable at \( x = -2 \) are:

\[
a = 92, \quad b = 142
\]

Would you like further clarification on any part of the solution? 

Here are 5 questions to expand on this problem:
1. How would the process change if the continuity point was at \( x = 1 \)?
2. What if the derivative of the second part was quadratic instead of linear?
3. Why does ensuring differentiability imply continuity, but not the reverse?
4. How would this problem change if there was a discontinuity at \( x = -2 \)?
5. What are the geometric implications of this function being continuous and differentiable at \( x = -2 \)?

**Tip**: When solving for continuity and differentiability at specific points, always start by equating the values of the function and its derivative at that point.

Find values of a and b such that a given piecewise function is both continuous and differentiable at a certain point.

Learn how to solve for constants a and b to ensure that a given piecewise function is both continuous and differentiable at a specific point. Suitable for undergraduate calculus students.

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