Graphs change from increasing to decreasing at points where the
tangent line is horizontal (has slope 0)
So we find the points on the graph where the tangent line is
horizontal. The derivative is a formula for the slope of a
tangent line, so we find the derivative and set it equal to
0, and solve for the values of x when the graph has a
horizontal tangent line:
f(x) = 2x³ + 2x² - 2x + 2
f'(x) = 6x² + 4x - 2 <--derivative, set = 0
6x² + 4x - 2 = 0
Divide through by 2
3x² + 2x - 1 = 0
Factor:
(x + 1)(3x - 1) = 0
x + 1 = 0; 3x - 1 = 0
x = -1; 3x = 1
x = 1/3
We put those points on a number line:
------o-------o------
-1 1/3
That determines three intervals, left of -1,
between -1 and 1/3, and right of 1/3, which
in interval notation are (-oo,-1), (-1,1/3), (1/3,oo).
We do a first derivative test on each interval
to determine whether the slope of the tangent line
is positive or negative. It is increasing if the
derivative, (slope of tangent line) is +, and
decreasing if it is -.
We make this chart to fill in, for each of the three
intervals:
Interval: (-oo,-1) | (-1,1/3) | (1/3,oo) |
Test pt.: | | |
Sign of f' | | |
incr or decr? | | |
For the test points, we choose any number on the
interval, and we may as well choose the easiest
one to substitute. On the interval (-oo,-1), we
will choose x=-2. On the interval (-1,1/3), we will
choose x=0. On the interval (1/3,oo), we will
choose x=1:
Interval: (-oo,-1) | (-1,1/3) | (1/3,oo) |
Test pt.: -2 | 0 | 1 |
Sign of f' | | |
incr or decr? | | |
For the sign of f', we substitute each of those values
in the equation for f'(x)
f'(x) = 6x² + 4x - 2
f'(-2) = 6(-2)² + 4(-2) - 2
f'(-2) = 6(4) - 8 - 2
f'(-2) = 24 - 8 - 2
f'(-2) = 14, which is + so we put a + for the sign of f'
and since it is +, it is increasing, so we write "incr"
on the bottom line.
f'(x) = 6x² + 4x - 2
f'(0) = 6(0)² + 4(0) - 2
f'(0) = 0 + 0 - 2
f'(-2) = -2, which is - so we put a - for the sign of f'
and since it is -, it is decreasing, so we write "decr"
on the bottom line.
f'(x) = 6x² + 4x - 2
f'(1) = 6(1)² + 4(1) - 2
f'(1) = 6(1) + 4 - 2
f'(1) = 6 + 4 - 2
f'(1) = 8, which is + so we put a + for the sign of f'
and since it is +, it is increasing, so we write "incr"
on the bottom line. The chart looks like this:
Interval: (-oo,-1) | (-1,1/3) | (1/3,oo) |
Test pt.: -2 | 0 | 1 |
Sign of f' + | - | + |
incr or decr? incr | decr | incr |
At -1, the point between the first two intervals, the
graph changed from increasing to decreasing, so it
reached a relative maximum point where x = -1,
so the graph is increasing on the interval (-oo,-1).
At 0, the point between the last two intervals, the
graph changed from decreasing to increasing, so it
reached a relative minimum point where x = 1/3,
so the graph is decreasing on the interval (-1,1/3).
At 1/3, the point between the first two intervals, the
graph changed from increasing to decreasing, so it
reached a relative maximum point where x = 1/3,
so the graph is increasing on the interval (1/3,oo).
We check by drawing the graph:
The graph is going uphill to the right left of the
point where x=-1, which is the relative maximum
point (-1,4), so it is increasing on the interval
left of that point.
Then the graph goes downhill to the right between
the points where x=-1 and where x=1/3, which
are the points (-1,4) and (1/3,44/27), which is
a relative minimum point. So the graph is decreasing
on the interval between those two points.
Then after leaving the point (1/3,44/27) the graph
is going uphill to the right and thus is increasing
right of that point.
Edwin
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