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Extrema and Average Rates of Change (vocabulary (A function f is…
Extrema and Average Rates of Change
objectives
Determine intervals
on which functions
are increasing,
constant, or
decreasing, and
determine maxima
and minima of
functions.
Determine the
average rate of
change of a function.
vocabulary
A function f is
increasing
on an interval I
if and only if for any two points in I, a
positive change in x results in a positive
change in f ( x ).
A function f is
decreasing
on an interval I
if and only if for any two points in I, a
positive change in x results in a negative
change in f (x ).
A function f is
constant
on an interval I
if and only if for any two points in I , a
positive change in x results in a zero
change in f (x ).
Critical points
of a function are those points at which a line drawn tangent to the curve is horizontal or vertical
Extrema
are critical points at which a function changes its increasing or decreasing behavior.
maximum
or a
minimum
value, either relative or absolute.
A
point of inflection
can also be a critical point
The
average rate of change
between any two points on the graph
of f is the slope of the line through
those points.
The line through two points on a
curve is called a
secant line
. The
slope of the secant line is denoted m sec .
examples
1
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing,
decreasing, or constant
f (x) = -2 x 3
Wh en viewed from left to right, the graph of f falls
for all real values of x. Therefore, we can conjecture
that f is decreasing on (-∞, ∞).
2
Estimate and classify the extrema for the graph of f (x).
It appears that f (x) has a relative maximum at x = -0.5
and a relative minimum at x = 1. It also appears that
lim
x:-∞
f (x) = -∞ and lim
x:∞
f (x) = ∞, so we conjecture that
this function has no absolute extrema.
3
Find the average rate of change of f (x) = - x 3 + 3x on each interval.
a. [-2, -01]
Use the Slope Formula to find the average rate of change of f on the interval [-2, -1].(x2) - f (x1) (x2) - f (x1)
[-(-1 ) 3 + 3(-1)] - [-(-2 ) 3 + 3(-2)]
.= -4
