To solve this problem, we need to follow these steps:

Part (a): Find the linear approximation $L(x)$ at $x = 3$

The linear approximation of a function $f(x)$ at a point $x = a$ is given by: $L(x) = f(a) + f'(a)(x - a)$

First, we need to find $f(3)$: $f(x) = 2x^2 - 5x + 5$ $f(3) = 2(3)^2 - 5(3) + 5$ $f(3) = 2(9) - 15 + 5$ $f(3) = 18 - 15 + 5 = 8$

Next, we need to find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(2x^2 - 5x + 5)$ $f'(x) = 4x - 5$

Then, find $f'(3)$: $f'(3) = 4(3) - 5$ $f'(3) = 12 - 5 = 7$

Now, we can write the linear approximation $L(x)$ at $x = 3$: $L(x) = f(3) + f'(3)(x - 3)$ $L(x) = 8 + 7(x - 3)$ $L(x) = 8 + 7x - 21$ $L(x) = 7x - 13$

Part (b): Use the linear approximation to approximate $f(3.2)$

Using the linear approximation $L(x) = 7x - 13$: $L(3.2) = 7(3.2) - 13$ $L(3.2) = 22.4 - 13$ $L(3.2) = 9.4$

Therefore, the approximation of $f(3.2)$ using the linear approximation is 9.4.

Would you like further details or have any questions?

Here are some related questions you might be interested in:

1. How do you find the second derivative of a function?
2. What is the purpose of linear approximation in calculus?
3. Can you explain the concept of Taylor series?
4. How do you determine the concavity of a function?
5. What are some real-world applications of linear approximations?

Tip: When approximating functions, the closer your point of approximation is to the given point, the more accurate your approximation will be.