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To solve this problem, we need to determine the slope (krypties koeficientas) of the tangent line to the function \( f(x) = x^4 - 2x^3 - 2 \) at the point where \( x = 4 \).

The slope of the tangent line at any point \( x = x_0 \) is given by the derivative \( f'(x) \) evaluated at \( x = x_0 \).

### Step 1: Differentiate \( f(x) \)
Given:
\[
f(x) = x^4 - 2x^3 - 2
\]

The derivative \( f'(x) \) is:
\[
f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x^3) - \frac{d}{dx}(2)
\]
\[
f'(x) = 4x^3 - 6x^2
\]

### Step 2: Substitute \( x = 4 \) into \( f'(x) \)
Now, we calculate \( f'(4) \):
\[
f'(4) = 4(4)^3 - 6(4)^2
\]
\[
f'(4) = 4 \cdot 64 - 6 \cdot 16
\]
\[
f'(4) = 256 - 96 = 160
\]

### Conclusion
The slope of the tangent line at \( x = 4 \) is 160, which corresponds to option **a**.

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. How would the solution change if the function were \( f(x) = x^4 - 3x^3 - 2 \)?
2. What is the geometric interpretation of the tangent line’s slope at a given point?
3. How can you find the equation of the tangent line at \( x = 4 \) for this function?
4. What is the impact of each term in \( f(x) = x^4 - 2x^3 - 2 \) on the function’s shape?
5. How does the second derivative \( f''(x) \) relate to the concavity at \( x = 4 \)?

**Tip:** When differentiating polynomials, remember that each term can be handled independently before combining the results.

Apskaičiuokite funkcijos f(x) = x^4 - 2x^3 - 2 grafiko liestinės, nubrėžtos per tašką, kurio abscisė x₀ = 4, krypties koeficientą.

Learn how to calculate the slope of the tangent line to the function f(x) = x^4 - 2x^3 - 2 at x = 4. This calculus problem involves taking derivatives and applying the power rule, suitable for high school students in Grades 11-12.

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