Here's the solution to the problem: Given the function $f(x) = 4x - 1$. a) Find the value of $f(2.999)$ and $f(3.001)$. Step 1: Calculate $f(2.999)$. Substitute $x = 2.999$ into the function: $$f(2.999) = 4(2.999) - 1$$ $$f(2.999) = 11.996 - 1$$ $$f(2.999) = 10.996$$ Step 2: Calculate $f(3.001)$. Substitute $x = 3.001$ into the function: $$f(3.001) = 4(3.001) - 1$$ $$f(3.001) = 12.004 - 1$$ $$f(3.001) = 11.004$$ The values are $f(2.999) = \boxed{10.996}$ and $f(3.001) = \boxed{11.004}$. b) Is function $f(x)$ continuous at the point $x=3$? Give reason. Yes, the function $f(x) = 4x - 1$ is continuous at the point $x=3$. Reason: A function $f(x)$ is continuous at a point $x=c$ if three conditions are met: 1. $f(c)$ is defined. 2. $\lim_{x \to c} f(x)$ exists. 3. $\lim_{x \to c} f(x) = f(c)$. For $f(x) = 4x - 1$ at $x=3$: 1. $f(3) = 4(3) - 1 = 12 - 1 = 11$. So, $f(3)$ is defined. 2. $\lim_{x \to 3} (4x - 1) = 4(3) - 1 = 11$. So, the limit exists. 3. $\lim_{x \to 3} f(x) = 11$ and $f(3) = 11$. Thus, $\lim_{x \to 3} f(x) = f(3)$. Alternatively, and more concisely, $f(x) = 4x - 1$ is a polynomial function. All polynomial functions are continuous for all real numbers. Therefore, $f(x)$ is continuous at $x=3$.