Download Example: Find intervals where the function f(x) = x 2 − 16 x is

Mt 020.02
Slide 1 on 3/22/00
Example: Find intervals where the
16
2
function f(x ) = x −
is increasing,
x
where it is decreasing and find its
relative extreme values.
Solution: Make a sign chart for the
16
derivative function f ' (x ) = 2x + 2 . Use
x
your calculator to plot it:
The derivative, f’, is negative for x<-2,
positive for x>-2 and f’(-2)=0. Note f’
does not exist at x=0. Critical points are
x=-2,0, because f’(-2)=0 and f’(0) d.n.e.
raj
Mt 020.02
Slide 2 on 3/22/00
Here’s our sign chart for the derivative f’:
f’ -------|+++++++++|++++++++++++++
-2
0
Thus the original function, f(x), is
decreasing on the interval (-infinity,-2),
and increasing across (-2,0) and
(0,infinity). Since the sign of f’ changes
across x=-2 from neg. to pos. we see
that x=-2 is the location of a relative min.
of the function f(x), and that f(-2)=12 is
the relative min. Here’s the graph of f
and you can see the results are correct:
raj
Mt 020.02
Slide 3 on 3/22/00
Example: The concentration of a certain
drug in a patient’s blood t hours after
t 2
injection is C (t) =
for 0≤t≤4 and
3
2t +1
C is measured in milligrams per cubic
centimeter. When is the concentration
of the drug increasing and when is it
decreasing?
Solution: Differentiate C(t) and draw a
2t(1− t 3)
sign chart. C ' (t ) =
and we get
3
2
(2t +1)
f’ +++++++++|---------------------------------1
raj
Mt 020.02
Slide 4 on 3/22/00
This sign chart tells us that the
concentration of the drug in the blood
increases (f’>0) for 0≤t≤1, but decreases
thereafter (since f’<0 after t=1).
Concentration reaches a max at t=1.
We move forward to finding a way to
distinguish between two kinds of
increasing.
We call them concave up (holds water)
and concave down (spills water).
Observe what happens to the tangent
lines’ slopes as we cross these two
different curves with increasing x values.
raj
Mt 020.02
Slide 5 on 3/22/00
Across the concave up, the slopes go
from small+, to medium+ to large+,
obviously growing bigger, thus
increasing. The SLOPE is increasing as
we increase x, the DERIVATIVE is what
we called an increasing function, and
that is signaled by ITS derivative being
positive. So (f’)’ = f’’>0, the second
derivative being positive signals f’ is
increasing which says that f, itself, the
original function is concave up. A
similar argument says f’’<0 forces f to be
concave down. Pictures below:
raj
Mt 020.02
Slide 6 on 3/22/00
Example: Find all interesting features of
4 3
5
the function f(x ) = 0.05x − x + 3x − 7.
3
In particular find where it is increasing,
decreasing, concave up, concave down,
relative extremes and where the
convavity changes (points of inflection).
Solution: We’ll need graphs and sign
charts for both f’ and f’’ to solve this.
Here are both for f’:
f’ +++++|------------|++++|--------|++++++
-3.9
-.89 .89
3.9
raj
Mt 020.02
Slide 7 on 3/22/00
So f is increasing on (-inf,-3.9),(-.89,.89)
and (3.9,inf) and decreasing on
(-3.9,-.89) and (.89,3.9). These x-values
can be obtained using the ‘intersect’
feature of your calculator with y=0 and
y=the derivative function. It looks like f
takes local max values at x=–3.9, and
x=.89 and local mins at x=-.89 and
x=3.9.
And here are the graph and sign chart
for the second derivative, f’’
And the sign chart for f’’ is
raj
Mt 020.02
Slide 8 on 3/22/00
f’’ -----------|++++++++|-----------|+++++++
-2.83
0
2.83
This chart speaks about the convavity of
the graph of f. It seems that the graph
of f is concave down across (-inf, -2.83)
and also across (0,2.83). Also f is
concave up for x values in (-2.83,0), and
also across (2.83,inf).
Points of inflection (concavity changes)
occur at x=-2.83, x=0 and x=2.83.
Here’s a graph of f(x):
raj