Given the functions $f(x) = 3x - 1$ and $g(x) = 2x$, let's solve the tasks step by step:

a) $f(x)$

The function $f(x)$ is already given as: $f(x) = 3x - 1$ There is nothing more to calculate here since it's already defined.

b) $g'(x)$

To find $g'(x)$, we need to differentiate $g(x) = 2x$ with respect to $x$. The derivative of $2x$ is: $g'(x) = \frac{d}{dx}[2x] = 2$

e) $(f \circ g)'(x)$ or $(fog)'(x)$

This asks for the derivative of the composite function $f(g(x))$.

First, find the composition $f(g(x))$: $f(g(x)) = f(2x) = 3(2x) - 1 = 6x - 1$

Now, differentiate $f(g(x))$ with respect to $x$: $(f \circ g)'(x) = \frac{d}{dx}[6x - 1] = 6$

Summary:

• a) $f(x) = 3x - 1$
• b) $g'(x) = 2$
• e) $(f \circ g)'(x) = 6$

Would you like further clarification on any step?

Here are 5 related questions to extend your understanding:

1. What is the derivative of the function $h(x) = 3x^2 - 5x + 4$?
2. How do you find the derivative of a product of two functions?
3. How would the composition $(f \circ g)(x)$ change if $f(x) = x^2 + 1$?
4. What is the chain rule and when is it applied?
5. Can you find $g'(x)$ if $g(x) = x^3 - 4x^2 + 7x$?

Tip: When differentiating composite functions, always remember to apply the chain rule!