Question

Ap Calculus BC 2009 Question 5?

Answer 1

a) `f'(4)` will be given by the slope between `3` and `5`.

`f'(4) = (f(5) - f(3))/(5 - 3) = (-2 - 4)/2 = -3`

Therefore `f'(4) = -3`.

b) We have:

`=int_2^13 3dx - 5int_2^13f'(x)dx`

`=[3x]_2^13 - 5(f(13) - f(2))`

`=39 - 6 - (5(6 - 1))`

`=33 - 25`

`=8`

c) Imagine drawing rectangles under the graph of `f(x)` and adding up their areas. This is a riemann sum. A left riemann sum is when you use the values of `f(x)` on the left.

`int_2^13 f(x) dx = 1(1) + 2(4) + 3(-2) +5(3) = 18`

d) We're given that `f'(5) = 3`, os the slope of the tangent at `x= 5` is `3`.

`y - (-2) = 3(x - 5)`

`y + 2 = 3x - 15`

`y = 3x - 17`

Using this we see that

`y(7) = 3(7) - 17 = 4`

This is the maximum possible value since the graph is concave down at that point and therefore the tangent line lies above the graph. Therefore, `f(7) ≤ 4`, as required.

The secant line's slope is given by finding the slope between `x= 5` and `x= 8`.

`m = (3 - (-2))/3 = 5/3`

Thus the equation of the secant line is

`y -3 = 5/3(x- 8)`

`y =5/3x - 40/3 + 3 =5/3x - 31/3`

At `x =7` this has value

`y(4) = 5/3(7) - 31/3 = 4/3`

Since the function is increasing on `[5, 8]`, this is the minimum possible value for `f(7)`. Thus `f(7) ≥ 4/3`.

Hopefully this helps!