How to Use Riemann Sums and Integrals to Estimate and Evaluate Functions

gitty_678
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Homework Statement


(x, f(x))
(2,1)
(3,4)
(5,-2)
(8,3)
(13,6)

A) Estimate f '(4). Show work.
B) Evaluate the Intergral from 2 to 13 of (3 - 5f '(x))dx. show work
C) Use left riemann sum with subintervals indicated bye the data in the table to apporoximate the intergral from 2 to 13 of (f '(x))dx. show work.
D) Suppose f '(5) = 3 and f ''(x) <0 for all x in the closed interval 5 <or= x <or= 8 to show graph of f at x=5 to show that f(7) <or= 4. use the secant lin for the graph of f on 5 <or= x <or= 8 to show that f(7) >or= 4/3.

The Attempt at a Solution


I have no idea where to even begin... if you could just give me a starting point or an equation or something to start with that would be amazing!
 
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There's not really enough information to say for sure how to proceed. Are you given any information regarding f&#039;(x)? The answers to these questions will vary substantially depending on how you choose to interpolate between the given points.
 
A. on the AP test, it's sufficient to just find the slope by m = f(b)-f(a) / b-a
B. integrating (3- 5f'(x))dx gives 3x - 5f(x) + C, and by fundamental theorem of calculus you can now plug in 13 and 2
C. just a left riemann sum 1*f(2) + 2*f(3) + 3*f(5) + 5*f(8)
D. on the actual exam it also mentioned being twice differentiable and says something about being concave down. from here, you can draw conclusions based on tangent line approximation, secant line approximation, and common sense with concavity.
 
oh yeah i forgot to put this

Let f be a function that is twice differentiable for all real numbers. The table above gives values of f for selected ponts on the closed interval 2 <or= x <or= 13

and yes this is from the AP test... my teacher is making us do all of them from the test.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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