---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- # We see, in detail, an exercise that is not easy to solve by hand (not even with # WoframAlpha, which, like much of the symbolic computing software, is not reliable for # the study of functions: it is necessary to use tricks to work withodd exponent powers# and to avoid making illicit simplifications of the type, for instance, √-1/√-1 -> 1). # # Study the graph of x →(x/4)and identify the # (local) minimum and maximum points and the concavity changes # (i.e. the inflection points) # # First of all I find the problem that for software z^a is defined only for z ≥ 0; then # I use rad3 that in this version of R is defined everywhere:^{2/3}- (x-4)^{1/3}g = function(x) rad3((x/4)^2) - rad3(x-4) # # First I study the function graphically, without "calculations". # I start with a wide interval, and then narrow it.graphF( g, -100,100, 1)# I get thefirst graphbelow to the left. # To confirm that for x that grows or decreases g(x) tends to grow beyond any limit, we # make numerical tests (try with 10, 100, 1000, ... and -10, -100, -1000, ..., and round # the outputs to shorten them)n=1:12; round(g(10^n))#[1] 0 4 30 163 809 3869 18205 85035 395850 1839861 8545238 39675026 n=1:12; round(g(-10^n))#[1] 4 13 50 206 901 4069 18636 85963 397850 1844170 8554521 39695026# I see that in both cases the outputs multiply (about) by 100 when the input is multi- # plied by 1000. OK [I can say it grows like |x|^(2/3), which is understandable by the # expression of g(x) since 2^3 > 1/3 (see at the bottom the graphs in a wider range)]. # Formally, thelimitfor x -> Inf and for x -> -Inf is Inf. # So I can narrow the input interval:graphF( g, -10,100, 1)# I get thesecond graph.# I have red marked those that at first sight might be the points sought. Let's first # look at the left side of the graph of g. graphF( g, -2,12, 1) # I get thegraphaboveto the right.# The program does the chart by pointing it: the dotted vertical line does not mean # that the function is not continuous. I can check this with some zoom: graphF( g, 3.9,4.1, 1)# I get the chart above to theleft. # # Looking at the expression of G, I perceive that for x=0 and x=4 (where x/4 and x-4 # change sign) I have critical points where there are a relative minimum and a # concavity change (inflection point). To study what happens to the right of 12 I will # need to calculate the derivatives. Let's see what we can do to the left without them. # h = minmax(g, -2,2); h; g(h)#[1] -1.589587e-14 "pratically" 0#[1] 1.587401 # I try to put 1.587401 in WolframAlpha, and I see it is an approximation of 2^(2/3) # I check this: g(0) = rad3(4), that is 4^(1/3), or (2^2)^(1/3) = 2^(2*1/3) = 2^/3) # SoA=(0,rad3(4))is arelative minimumpoint. # # For x=4 g(x)=1. As I understand from the graph inC=(4,1)I have a vertical # inflection point. # # Let's look at B and D: j = minmax(g, 1,3); j; g(j); k = minmax(g, 5,10); k; g(k)#[1] 2#[1] 1.889882#[1] 8#[1] 0 # If I put 1.889882 in WolframAlpha I see it is an approximation of 3/2^(2/3) # [=3*2^(1/3)/2]. I can (as above) check this.B=(2,3*rad3(2)/2)is arelative#maximumpoint. InsteadD=(8,0)is arelative minimumpoint. # # Let's study the part of the graph to the right of point D. # Is there an oblique inflection point? Above I marked a hypothetical point E of that # kind. To test this I have to calculate the first derivative first and see where it # grows or decreases (orI have to calculate the second derivative). I do both with R. # I use this also to study the problems that I have solved earlier. # # I can calcolate g' with deriv(g,"x"); I can save it as function with # dfg <- function(x) eval(deriv(g,"x"))deriv(g,"x")#Error: ...# The definition of g contains rad3 which is defined as follows: rad3#function(x) sign(x)*abs(x)^(1/3) # I should break the definition of g right and left of 4, where the input of rad3(x-4) # changes sign.g1 = function(x) ((x/4)^2)^(1/3)+(4-x)^(1/3) g2 = function(x) ((x/4)^2)^(1/3)-(x-4)^(1/3) # I check the thing graphically (graph above to theright):HF=2.7; BF=2.7 graphF( g, -2,12, 1) graph( g1, -2,4, "red") graph( g2, 4,12, "blue") # OK dg1 <- function(x) eval( deriv(g1,"x") ) solution(dg1,0, 1,3)#[1] 2 dg2 <- function(x) eval( deriv(g2,"x") ) solution(dg2,0, 5,10)#[1] 8 # I foundBandDagain. But are they the same values? more( minmax(g, 1,3) ); more(solution(dg1,0, 1,3))#1.99999998773738 2 # The second method is more accurate because the algorithm to find the minimum and # maximum without the derivative (although being more general) is more approximate # (but differences are generally negligible from the point of view of application). # # To findEI can study where grow/decrease D(g) maxmin(dg2,10,100)#[1] 31.52936# Or I can study where the 2nd derivative is 0 d2g2 <- function(x) eval( deriv2(g2,"x") ) solution(d2g2,0, 10,100)#[1] 31.52936# I found the same value. In reality, the two procedures do not give the same values: more( maxmin(dg2,10,100) ); more( solution(d2g2,0, 10,100) )#31.5293616373527 31.5293611968877# Again, thesecondmethod gives a more accurate result. # The corresponding y is: g( solution(d2g2,0, 10,100) )#[1] 0.9412041 # SoE=(31.52936, 0.9412041)# # If I want I can calculate the derivatives by hand, or with R. By deriv(g2,"x") I get # ((x/4)^2)^((1/3)-1)*((1/3)*(2*(1/4*(x/4))))-(x-4)^((1/3)-1)*(1/3) which I can # simplify in (x/4)^(-1/3)/6 - (x-4)^(-2/3)/3 ... or I can use WolgramAlpha. If # I put: solve ( d/dx d/dx ((x/4)^2)^(1/3)-(x-4)^(1/3) ) = 0 # I get: 31.52936119688764498693379… # # Here's how thegraphwith the solutions was obtained:HF=2.7; BF=2.7*2 graphF( g, -1,40, 1) graph ( g, 3,5, 1) a = 0; b = 2; c = 4; d = 8; e = solution(d2g2,0, 10,100) POINT( c(a,b,c,d,e), g( c(a,b,c,d,e) ), "red")# other elements on the chart:coldash="red"; gridV(2); gridV(4); gridV(8) type(0,1.4,"A"); type(2,2.1,"B"); type(5.5,1,"C") type(8,0.2,"D"); type(31.53,0.75,"E") # # - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - # Comparison of g with: f = function(x) abs(x)^(2/3) graphF (g, -1e5,1e5, "black"); graphF (f, -1e5,1e5, "brown")# I see that (for x that tends to Inf and to -Inf) both curves rise beyond any # limitations and keep concavity downOther examples of use