<p>Right now I have a TI 89. I've had it for the past 2 years and sadly I'm clueless [not entirely] about using the magnificent functions of it.</p>
<p>Can anyone show my the calculations that would help best when taking the Math2C test, such as minimum and maximum?</p>
<p>Oh but one thing, i know how to use zeros() and solve() :]</p>
<p>Well, for max and min just go to the Homescreen (the screen where you actually can calculate stuff), then press F3 and go to fmax or fmin. </p>
<p>Then you see something like: fMin(</p>
<p>now type in your function and the respective variable.</p>
<p>Example: fMin(x^2,x) ---> x=0</p>
<p>I never really remembered how to use the fMIn and fMax. Haha. I would just differentiate it (under calculus) and get the function, then solve for the zero of the differentiated function. For example, finding the minimum of (sin x)/x,
d(sinx/x, x) <em>enter</em> = cos x/x - sinx/x^2
solve ((cos x/x - sin x/x^2) = 0, x) <em>enter</em> x = Bla Bla.
Actually, I didn't really rely on my TI so, it's possible to do well without being familiar with the functions. Just don't forget where pi is on the keyboard and spend 5 minutes looking for it... :eyeroll:</p>
<p>Well, I really don't want to advocate my approach to the task of finding a minimum or maximum :P but I have to mention that your described approach may have some flaws for some exercises. </p>
<p>The condition for a min/max is f'(x)=0 AND f''(x) not 0. So first you actually ought to check whether the x coordinate you got out the f'(x) is zero or not; if not, this point wouldn't be a max/min. </p>
<p>I would recommend my method. </p>
<p>Sorry, fiona! ;)</p>
<p>Thanks a lot you guys! Actually, I'm in AP Calculus BC hehe, but i didn't want to mix up calculus with these non-calculus problems. but thanks! really, doing it by hand should be better.</p>
<p>"Well, I really don't want to advocate my approach to the task of finding a minimum or maximum :P but I have to mention that your described approach may have some flaws for some exercises.</p>
<p>The condition for a min/max is f'(x)=0 AND f''(x) not 0. So first you actually ought to check whether the x coordinate you got out the f'(x) is zero or not; if not, this point wouldn't be a max/min.</p>
<p>I would recommend my method.</p>
<p>Sorry, fiona!"</p>
<p>But can't you just check that on a number line, to see whether there is a change in the rate of change and thus, whether it is a max or a min or neither?</p>
<p>But I still advocate your method because, well, this IS a timed test, and calculus can take a while. XD</p>
<p>Gosh, what have I written? This was partially ******** ^^.</p>
<p>You ought to check whether the x-coordinate which you got outta f'(x)=0 is zero for f''(x); if not, it IS of course a min/max. Sorry about that. (because the second condition is f''(x) NOT 0).</p>
<p>Boys and girls, good luck tomorrow!</p>