To solve this question using linear approximation, we use the following formula:

$L(x) = f(a) + f'(a)(x - a)$

Here:

• $a = 2$ (a nearby point),
• $f(2) = 8$ (function value at $x = 2$),
• $x = 2.1$ (point to approximate),
• $f'(x) = -\frac{3x^2}{\sqrt{x^3} + 1}$ (given derivative).

Step 1: Calculate $f'(2)$

Substitute $x = 2$ into $f'(x)$:

$f'(2) = -\frac{3(2)^2}{\sqrt{(2)^3} + 1}.$

Simplify each component:

• $2^2 = 4$, so $3(2)^2 = 3(4) = 12$,
• $2^3 = 8$, so $\sqrt{2^3} = \sqrt{8} = 2\sqrt{2}$.

Thus:

$f'(2) = -\frac{12}{2\sqrt{2} + 1}.$

Step 2: Approximate $f(2.1)$ using linearization

Now use the linear approximation formula $L(x) = f(a) + f'(a)(x - a)$. Substituting:

• $f(a) = 8$,
• $f'(2) = -\frac{12}{2\sqrt{2} + 1}$ (from Step 1),
• $x - a = 2.1 - 2 = 0.1$.

The formula becomes:

$f(2.1) \approx 8 + \left( -\frac{12}{2\sqrt{2} + 1} \right)(0.1).$

Simplify the result numerically (to 2-3 decimal places, as needed).

Would you like me to evaluate the numeric result completely? Let me know if further clarifications are needed!

Related Questions:

1. What is the concept of linear approximation and its applications?
2. How can you approximate values of functions using derivatives?
3. What happens if you increase $h$ (distance from $a$) in linear approximation?
4. How does the behavior of $f'(x)$ impact the accuracy of the approximation?
5. Can you extend this idea to quadratic approximations?

Tip: Linear approximation works well for small changes around a given point, especially when the function is smooth and differentiable.