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To find the values of \( a \) and \( b \) such that the function

\[
f(x) =
\begin{cases} 
5x^3 - 6x^2 + 2, & x < -2 \\
ax + b, & x \geq -2 
\end{cases}
\]

is both continuous and differentiable at \( x = -2 \), we need to ensure the following conditions:

### 1. **Continuity at \( x = -2 \)**:
The left-hand and right-hand limits of the function must be equal at \( x = -2 \), meaning:
\[
\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x)
\]
This implies:
\[
5(-2)^3 - 6(-2)^2 + 2 = a(-2) + b
\]
Simplifying:
\[
5(-8) - 6(4) + 2 = -2a + b
\]
\[
-40 - 24 + 2 = -2a + b
\]
\[
-62 = -2a + b \tag{1}
\]

### 2. **Differentiability at \( x = -2 \)**:
The derivatives from the left-hand and right-hand sides must be equal at \( x = -2 \), meaning:
\[
\lim_{x \to -2^-} f'(x) = \lim_{x \to -2^+} f'(x)
\]
First, we calculate the derivative of each piece of the function.
- For \( x < -2 \), \( f(x) = 5x^3 - 6x^2 + 2 \). Differentiating:
  \[
  f'(x) = 15x^2 - 12x
  \]
  So at \( x = -2 \):
  \[
  f'(-2) = 15(-2)^2 - 12(-2) = 15(4) + 24 = 60 + 24 = 84
  \]
- For \( x \geq -2 \), \( f(x) = ax + b \). Differentiating:
  \[
  f'(x) = a
  \]
  Thus, for differentiability:
  \[
  f'(-2) = a = 84 \tag{2}
  \]

### 3. **Solving for \( a \) and \( b \)**:
From equation (2), we already have \( a = 84 \).

Substitute \( a = 84 \) into equation (1):
\[
-62 = -2(84) + b
\]
\[
-62 = -168 + b
\]
\[
b = -62 + 168 = 106
\]

Thus, the values are:
\[
a = 84, \quad b = 106
\]

Would you like more details on any step? Here are 5 related questions to deepen your understanding:

1. How can we generalize this method for any piecewise function to ensure continuity?
2. Why is ensuring differentiability important in applications like physics?
3. What happens if only continuity, but not differentiability, is required for a function?
4. How would the process change if we were given a function with more than two pieces?
5. What does it mean for a function to be differentiable at a point geometrically?

**Tip**: Always check continuity before differentiability when working with piecewise functions!

Find a and b so that the function f(x) = { 5x^3 - 6x^2 + 2, x < -2; ax + b, x ≥ -2 } is both continuous and differentiable.

This problem demonstrates how to find the values of a and b such that a piecewise function is both continuous and differentiable. It explores key concepts such as limits, continuity, and differentiability. Suitable for undergraduate calculus students.

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