Finding the Maximum/Minimum value of a function

This seems to be quite a simple thing to do, but for some ungodly reason i am having unimaginable trouble doing so, on my Casio calculator its simple as Pie, you put in the function and say find maximum between this point and this point, however I cant for the life of me figure out how to do it on this, and i tried a whole bunch of things that kept giving me errors, i was hoping if someone could help me with this it will be amazing.

What i would like to know is.... How to find the maximum of a function, I am not posting the whole code, just the thing that i am having trouble with..... and this is it....

x(t)=0.03106*exp(-21.43*t)*sin(32.2*t)
v(t)=diff(x(t));
a(t)=diff(v(t));
jerk(t)=diff(a(t));
 G=solve(jerk(t)==0,t>0)
G=vpa(G,8)
 ezplot (a(t),[0,0.5]) ; title a(t)

Now this is where the problem begins... to solve for maximum acceleration I am differentiating acceleration and then making that equal to ZERO, which should give me the time at which acceleration is ZERO However It is outputting a value for G which is not coherent with the ezplot, why? and how do I fix this?

Any help would be very greatfully appreciated and thanks in advance!!!

Just a little bit more info, It seems that whatever is "outisde" the ezplot the equation COMPLETELY diregards, it is wierd but i am telling the truth, at one point i wanted to find the maximum of x(t), which say it was at t=0.5 but it showed me the maximum to be t=0.46 JUST because the top of the graph was cut and 0.46 was the "maximum" of the graph that ezplot saw what is going on???? if someone can please confirm I am not insane that would be great.

[Walter Roberson]

Is it correct that before this you have defined

syms x(t) v(t) a(t) jerk(t)

[Eli Kroitor]

i havent before but i have now and it didnt make a difference

Answer 1

[Walter Roberson]

What value of G is it outputting?

When you apply solve() to something that is not a polynomial, it is not certain that it will be able to find all of the roots; it might return only one numeric root and that one root could be anywhere. There are an infinite number of solutions spaced pi/32.2 apart. The first positive one is at approximately -(5/161)*arctan(10976809340/56816825393)+5/161*pi -- approximately because it depends on whether 0.03106 means 3106/10000 exactly or if the 0.03106 is to be interpreted as 0.03106 +/- 0.000005 .

[Eli Kroitor]

hey walter thanks for answering back, sorry for such a late reply, so here is my new code (i just added 0.15>t>0) which now "should" give me the maximum value, instead its saying something else.....

x(t)=0.03106*exp(-21.43*t)*sin(32.2*t)
v(t)=diff(x(t));
a(t)=diff(v(t));
jerk(t)=diff(a(t));
 G=solve(jerk(t)==0,0.15>t>0)
G=vpa(G,8)
 ezplot (a(t),[0,0.5]) ; title a(t)

The answer that G gives me now is as follows.... G = Empty sym: 0-by-1

[Walter Roberson]

As you are going to name a numeric range:

G = vpasolve(jerk(t)=0, t, [0, 0.15])

0.15>t>0 does not mean what you would expect: it is parsed as ((0.15>t)>0) and the first part (0.15>t) returns a value of 1 if the condition is true and 0 if the condition is false. So you would then proceed to test either 1>0 or 0>0 . The first of those is of course true and the second is false -- which turns out to be the same as if you had just tested 0.15>t . In MATLAB break up your range tests

 0.15 > 0 & t > 0

[Eli Kroitor]

Hey walter thank you so much for that! I now understand what i did wrong, i never even knew vpasolve even exsisted! again thank you so much, i always seem to have trouble with the simplest things in matlab, I think something should be straight forward and easy awhich leads me to cutting corners, and when it doesent work i just get more and more impatient which just leads to more and more mistakes, Thanks again haha