To solve this problem, we are interpreting the graph of $f'(x)$ to determine the values of $f(x)$ at specific points. The integral (area under the curve) of $f'(x)$ represents the change in $f(x)$, as given by the Fundamental Theorem of Calculus:

$f(b) = f(a) + \int_a^b f'(x) \, dx$

Step-by-step Explanation:

1. Starting Point: $f(8) = 4$ The value $f(8) = 4$ is provided as a reference point. We use this value to calculate $f(x)$ for other values of $x$.

2. a) $f(6)$ To find $f(6)$, we integrate $f'(x)$ from 8 to 6: $f(6) = f(8) + \int_8^6 f'(x) \, dx$
   • The integral from $x = 6$ to $x = 8$ corresponds to Area 7 in the graph, which is $-7$ (negative because it is below the x-axis). $f(6) = 4 + (-7) = -3$

3. b) $f(4)$ To find $f(4)$, integrate $f'(x)$ from 8 to 4: $f(4) = f(8) + \int_8^4 f'(x) \, dx$
   • The integral from $x = 4$ to $x = 8$ is the sum of Area 7 and Area 3:
     • Area 7 = $-7$
     • Area 3 = $+3$ $\int_8^4 f'(x) \, dx = -7 + 3 = -4$ $f(4) = 4 + (-4) = 0$

4. c) $f(1)$ To find $f(1)$, integrate $f'(x)$ from 8 to 1: $f(1) = f(8) + \int_8^1 f'(x) \, dx$
   • The integral from $x = 1$ to $x = 8$ is the sum of all areas:
     • Area 7 = $-7$
     • Area 3 = $+3$
     • Area 6 = $+6$
     • Area 2 = $-2$ $\int_8^1 f'(x) \, dx = -7 + 3 + 6 - 2 = 0$ $f(1) = 4 + (0) = -6$

5. d) $f(0)$ To find $f(0)$, integrate $f'(x)$ from 8 to 0: $f(0) = f(8) + \int_8^0 f'(x) \, dx$
   • Extend the integral to include all areas from $x = 0$ to $x = 8$:
     • Area 7 = $-7$
     • Area 3 = $+3$
     • Area 6 = $+6$
     • Area 2 = $-2$
     • Area 8 = \ --Area Counts x!!