Let's go through each of the function evaluation problems in the image and solve them step by step:

1. Given the function $f(x) = 3x - 5$, find $f(4)$:

Substitute $x = 4$ into $f(x)$: $f(4) = 3(4) - 5 = 12 - 5 = 7$

2. Find the value of $h(b) = 3b^2 - 2b + 1$ when $b = -3$:

Substitute $b = -3$ into $h(b)$: $h(-3) = 3(-3)^2 - 2(-3) + 1 = 3(9) + 6 + 1 = 27 + 6 + 1 = 34$

3. Find $g(2w)$ when $g(x) = x^2 - 2x + 1$:

Substitute $x = 2w$ into $g(x)$: $g(2w) = (2w)^2 - 2(2w) + 1 = 4w^2 - 4w + 1$

4. Given $f(x) = 2x^2 + 4x - 3$, find $f(2a + 3)$:

Substitute $x = 2a + 3$ into $f(x)$: $f(2a + 3) = 2(2a + 3)^2 + 4(2a + 3) - 3$ First, expand $(2a + 3)^2$: $(2a + 3)^2 = 4a^2 + 12a + 9$ Now, substitute: $f(2a + 3) = 2(4a^2 + 12a + 9) + 4(2a + 3) - 3$ $= 8a^2 + 24a + 18 + 8a + 12 - 3$ $= 8a^2 + 32a + 27$

5. Given $f(x) = x^2 - 4$ and $g(x) = x + 1$, evaluate:

• a. $f(-2)$: $f(-2) = (-2)^2 - 4 = 4 - 4 = 0$

• b. $g(3)$: $g(3) = 3 + 1 = 4$

• c. $(f - g)(-1)$: First, evaluate $f(-1)$ and $g(-1)$: $f(-1) = (-1)^2 - 4 = 1 - 4 = -3$ $g(-1) = -1 + 1 = 0$ Now, $(f - g)(-1) = f(-1) - g(-1) = -3 - 0 = -3$

• d. $f(g(-2))$: First, find $g(-2)$: $g(-2) = -2 + 1 = -1$ Now, substitute into $f(x)$: $f(-1) = (-1)^2 - 4 = 1 - 4 = -3$

Let me know if you'd like any of these steps detailed further or have any questions!

Related Questions:

1. How do you evaluate functions with two variables like $f(a, b)$?
2. What happens if the function has a piecewise definition?
3. Can functions involve trigonometric terms, and how would we evaluate those?
4. How do you determine if a function is injective or surjective?
5. What are composite functions, and how are they evaluated?

Tip:

When working with composite functions like $f(g(x))$, always solve the inside function first, then substitute into the outer function.