This code is intended to allow students to visualize the logic behind the mean value theorem. It was written for Political Science 508, a mandatory course for first-semester political science Ph.D. students at Emory University, by Jane Lawrence Sumner.

mean-value-theorem.R

#Understanding the Mean Value Theorem
#POLS 508, Fall 2015
#TA: Jane Lawrence Sumner


# The Mean Value Theorem states that if the function is continuous on the closed 
# interval and differentiable on the open interval, then there exists a point c 
# at which the tangent to that point is parallel to the secant.

## In other words, there exists a point c such that f'(c)=(f(b)-f(a))/(b-a)

## What does that look like? 


#### set your lower and upper bounds, function, and the derivative #####
a <- -10
b <- 10
x <- seq(a,b,by=.0001)
fx <- x^3
derivative <- 3*x^2

#### plots the function ####
plot(x,fx,type="l") 
fxb <- fx[which(x==b)]                                   #function evaluated at b
fxa <- fx[which(x==a)]                                   #function evaluated at a

slope.secant <- (fxb-fxa)/(b-a)                           #slope of the secant line
                                                          # this is what f'(c) must 
                                                          #equal
intercept.secant <- fxb-slope.secant*b                    #y-intercept (for plotting)
abline(coef=c(intercept.secant,slope.secant))             #visualize the secent


#at what point(s) c is the derivative of the function equal to the slope of the secant
#line? (note: I call this cpt instead of c, since c is "concatenate" in R. It is
#possible to rename R functions as variables, but it's not a great idea.)
# (note 2: if you're getting errors, the odds are good it has to do with the 
# rounding in the first line of cpt. As the secant slope becomes larger, the amount
# of rounding you should do decreases, so play around with this.)
cpt <- x[which(round(derivative,3) %in% round(slope.secant,3))] #first, locate them
cpt <- unique(round(cpt,2))                                 #because of how it matches
                                                            #to find x, there may be
                                                            #more than one that are
                                                            #approximately equal.
                                                            #this sorts that out.
cpt <- cpt[cpt %in% c(a,b)==0]                              #can't be a or b!


abline(v=cpt,lty=3)                                         #plot that.
fxc <- fx[which(round(x,4) %in% round(cpt,4))]                  #function evaluated at c
intercept.tangent <- fxc-slope.secant*cpt                   #just rearrange, since we 
                                                            #know the slope, x, and y
for(i in 1:length(intercept.tangent)){
abline(coef=c(intercept.tangent[i],rep(slope.secant,length(intercept.tangent))[i]))
}
# the above line of code just plots a line for as many lines as there should be
# because we have only one secant slope, but two lines, I just repeat the value
# for as many lines as we have