https://file.aiquickdraw.com/mathbot/file/1729539243533jeo2qfc4.jpg

bestem a og b slik at funksjonen er kontinuerlig og deriverbar.

For the function \( f(x) \) to be continuous and differentiable at \( x = 2 \), the following conditions must be met:

1. **Continuity at \( x = 2 \)**: The value of the function from both sides of \( x = 2 \) must be equal, i.e.,

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

2. **Differentiability at \( x = 2 \)**: The derivative from both sides at \( x = 2 \) must be equal, i.e.,

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

### Step 1: Continuity Condition
For continuity at \( x = 2 \), the two pieces of the function must be equal at \( x = 2 \).

\[
a \cdot 2^3 = 2 \cdot 2^2 + b.
\]

This simplifies to:

\[
8a = 8 + b.
\]

Thus, we get the first equation:

\[
8a - b = 8. \tag{1}
\]

### Step 2: Differentiability Condition
We need to compute the derivative of each piece of the function and then set them equal at \( x = 2 \).

- For \( x \leq 2 \), the derivative of \( f(x) = a \cdot x^3 \) is:

\[
f'(x) = 3a \cdot x^2.
\]

At \( x = 2 \):

\[
f'(2) = 3a \cdot 2^2 = 12a.
\]

- For \( x > 2 \), the derivative of \( f(x) = 2x^2 + b \) is:

\[
f'(x) = 4x.
\]

At \( x = 2 \):

\[
f'(2) = 4 \cdot 2 = 8.
\]

Now, for differentiability, these two derivatives must be equal:

\[
12a = 8.
\]

Solving for \( a \), we get:

\[
a = \frac{8}{12} = \frac{2}{3}. \tag{2}
\]

### Step 3: Solving for \( b \)
Substitute \( a = \frac{2}{3} \) into equation (1):

\[
8 \cdot \frac{2}{3} - b = 8,
\]

which simplifies to:

\[
\frac{16}{3} - b = 8.
\]

Multiplying everything by 3 to clear the fraction:

\[
16 - 3b = 24.
\]

Solving for \( b \):

\[
3b = 16 - 24 = -8,
\]

\[
b = -\frac{8}{3}. \tag{3}
\]

### Final Answer:
The values of \( a \) and \( b \) that make the function continuous and differentiable are:

\[
a = \frac{2}{3}, \quad b = -\frac{8}{3}.
\]

Let me know if you would like further details or clarification!

### Related Questions:
1. How would the function behave if the conditions were only for continuity, not differentiability?
2. What happens if we change the cubic term in the first part of the piecewise function?
3. Can you explain how differentiability implies continuity but not vice versa?
4. How do we find higher-order derivatives for piecewise functions like this one?
5. Could this method be applied if \( x = 2 \) were replaced by a different number?

### Tip:
For piecewise functions, always check both continuity and differentiability conditions separately. Continuity ensures the function is well-behaved, while differentiability guarantees smooth transitions.

Learn how to find constants a and b to ensure the function is continuous and differentiable at a point. This involves applying continuity and differentiability conditions to a cubic and quadratic function in a piecewise setup. Suitable for advanced high school or early university students.

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