Ap Calculus BC 2009 Question 5?

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!