Identify the steeper line y = 3x + 4 or y = 6x + 11
2 answers:
You might be interested in
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for 
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
, and substituting this into the recurrence, you find that 
for all 
.
Next, the linear term tells you that
, or 
.
Now, if
is the first term in the sequence, then by the recurrence you have
and so on, such that
for all 
.
Finally, the quadratic term gives
, or 
. Then by the recurrence,
and so on, such that
for all.
Now, the solution was proposed to be
so the general solution would be
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
Next, the linear term tells you that
Now, if
and so on, such that
Finally, the quadratic term gives
and so on, such that
for all
Now, the solution was proposed to be
so the general solution would be
Answer:
See Below.
Step-by-step explanation:
We are given the graph of <em>y</em> = f'(x) and we want to determine the characteristics of f(x).
Part A)
<em>f</em> is increasing whenever <em>f'</em> is positive and decreasing whenever <em>f'</em> is negative.
Hence, <em>f</em> is increasing for the interval:
And <em>f</em> is decreasing for the interval:
Part B)
<em>f</em> has a relative maximum at <em>x</em> = <em>c</em> if <em>f'</em> turns from positive to negative at <em>c</em> and a relative minimum if <em>f'</em> turns from negative to positive to negative at <em>c</em>.
<em>f'</em> turns from positive to negative at <em>x</em> = -2 and <em>x</em> = 1.
And <em>f'</em> turns from negative to positive at <em>x</em> = -1 and <em>x</em> = 3.
Hence, <em>f</em> has relative maximums at <em>x</em> = -2 and <em>x</em> = 1, and relative minimums at <em>x</em> = -1 and <em>x</em> = 3.
Part C)
<em>f</em> is concave up whenever <em>f''</em> is positive and concave down whenever <em>f''</em> is negative.
In other words, <em>f</em> is concave up whenever <em>f'</em> is increasing and concave down whenever <em>f'</em> is decreasing.
Hence, <em>f</em> is concave up for the interval (rounded to the nearest tenths):
And concave down for the interval:
Part D)
Points of inflections are where the concavity changes: that is, <em>f''</em> changes from either positive to negative or negative to positive.
In other words, <em>f </em>has an inflection point wherever <em>f'</em> has an extremum point.
<em>f'</em> has extrema at (approximately) <em>x</em> = -1.5, 0, and 2.2.
Hence, <em>f</em> has inflection points at <em>x</em> = -1.5, 0, and 2.2.