Calc II Question, please help!?

the graph of f is just a bunch of line segments that connects each other. As long as you know the value of one particular x value, you'll know the value of the functions at all x values since f is continuous


we know (3,0) on the graph. And so from 2 <= x <= 3, the portion of the graph is -1(x-3).

So, f(2) = -1(2-3) = 1.

from 0 <= x <= 2, the portion of the graph is 1(x-2) + 1.

you have f(0) = 1(0-2) + 1 = -1

You can use the same process to find the value of f(7), which turns out to be -1.

f(0) = -1 and f(7) = -1

b)
Find the areas by calculating the areas of several rectangles:
rectangle 1: (2)(1) = 2
rectangle 2 : (-1)(1) = -1
rectangle 3: (2)(1) = 2
rectangle 4: (-2)(2) = -4
rectangle 5: (1)(1) = 1

Add:
7
∫ f'(x) dx = 2-1+2-4+1 = 0
0

a)
Not sure of my answer:
Both f'(0) and f'(7) are undefined so f is not differentiable at x=0 and x=7 and cannot be continuous at x=0 and x=7.
But the question says f(x) is continuous.