1. F’ means derivative or in this case slope.
Part (a)
F’ means derivative or in this case slope.
From Algebra I, slope is rise over run. We will have to use this since we only have a chart. Use the points (3,4) and (5,-2)
M= (-2)-(4)
(5)-(3)
= -3

2. = 8 ∫ 3 - ∫ 5 f’(x) [3x] - 5[ f(x)] 3(13) - 3(2) - 5(6) –5(1) 33 - 25
Part (b)
Integrate parts separately
∫ ∫ 5 f’(x) 13 2
[3x] [ f(x)] 13 2
3(13) - 3(2) (6) –5(1)
= 8

3. = 18 Part (c) Left Riemann Sum
f(2)(3-2) + f(3)(5-3) + f(5)(8-5) + f(8)(13-8)
(1)(3-2) + (4)(5-3) + (-2)(8-5) + (3)(13-8)
1 + 8 –
= 18

4. Part (d)
Using a tangent line approximation we can make an equation with a slope of three going through (5,-2)
(y+2)= 3(x-5) Putting in a 7 for x yields the point (7,4)
Since the graph is concave down the tangent approximation line overestimates the value of the function so therefore f(7) ≤ 4
Using a secant line we find the slope to be 5/3!!!
3-(-2)
8-5
Using 5/3 as the slope we can then write another line equation (y+2)= 5/3 (x-5) so when x=7 f(x)=4/3
Since the graph is concave down, the secant line underestimates the values so 4/3 is less than the actual value at x=7 so therefore f(7) ≥ 4/3