.:. MORE ON DERIVATIVES .:.
W by Flickr user Leo Reynolds
We started off right away on the math by first plugging a function into our
calculators. The shape on the cover page is a quot;wlikequot; fcn known as a
QUARTIC function, just like the function here.
f (x) = 3x4 4x3 12x2 + x + 10
Our specific instructions were not to graph the function right away, but
instead, change the intervals of the graph pressing WINDOW on our
Function in interval 1 (cont'd...)
INFO about this function at this interval
This function has:
• a LOCAL MAXIMUM [also absolute]
• an ABSOLUTE MINIMUM
ABSOLUTE MIN/MAX VALUE: These are also known as EXTREME VALUES
[global minimum and global maximum]. These are the max or min y values that are
not exactly restricted by a range or an interval.
LOCAL MAX/MIN: Usually a quot;maxquot; or quot;minquot; y value that are usually also
critical points of a function located in the range.
The second interval was [1.2, 2.5] [25, 5]
The function in this interval has:
• a local minimum as well as an absolute min.
• an absolute max value.
Finally, the third interval at [0.5, 2.8] [25, 25]
Since the function is shown on a larger scale it shows there is:
• a local min at the same position
• an absolute max value.
**NOTE: These again, are not three different functions, but one function shown
in three ways and different intervals. This shows us that in certain intervals, only
some local and/or global y values are shown within that range.
With that said, the EXTREME VALUE THEOREM is applicable. This can occur
on any continuous function in a closed interval so that extreme values can be
.:. EXTRA NOTES .:.
ANOTHER PERSPECTIVE: a DERIVATIVE is
a function that measures the slopes of
tangent lines [changes in a function].
A derivative is also a limit, since the slopes are
constantly trying to approach a certain value.
The ROOTS of a derivative are where the
parent function has tangent lines with m = 0
.:. FIRST DERIVATIVE TEST .:.
ex. 2 Given a function of f (x) = x2 4x + 5, determine
whether or not there are local min and/or max values within
the interval of [1, 3]
f '(x) = 2x 4 The first thing we do is find the derivative
of the function.
**This function can be further
factored into f '(x) = 2 (x 2) so that
f '(2) = 2(2) 4 it is easier to find the root of the
= 4 4 derivative, which is where this
f '(2) = 0 tangent line has a slope of 0.
Subsequently, this point is also the
vertex of this function.
Now that we have the vertex, we have an idea of what this graph will
look like. We know that this parabolic function will open upwards
because the coefficient of the first term in the fcn is posve.
Since this function is within the domain of a closed interval,
according to the extreme value theorem, the global min and/or max
values can be found, simply by plugging the coordinates into the
f (2) = 22 4(2) + 5
= 4 8 + 5
f (2) = 1 These calculated values of course are the y
coordinates of the function where x is at a
f (1) = (1)2 4(1) + 5 certain point, such as endpoint 1 or 3.
= 1 + 4 + 5
f (1) = 10 Finding these values can also tell us where the
critical points are as well as what the question
f (3) = 32 4(3) + 5 was asking for; the local min and max values.
= 9 12 + 5
f (3) = 2
OTHER THINGS WE LEARNED TODAY:
It was clarified that the derivative of a function at a cusp is supposed to be
undefined. The calculator says otherwise.
REASON WHY: In order to determine derivatives, the calculator strictly
follows the rule known as the SYMMETRIC DIFFERENCE QUOTIENT;
terminology for finding the slope, since the equation of a slope is the
quotient of two differences.
Furthermore, what the calculator did was find the slopes of tangents coming
from both sides approaching a value, which is usually how a derivative works,
but the value being approached is infinity, which is not really a number, but a
concept of a number.