(No links) Tina's potpourri uses 15 drops of lavender oil for every 3 cups of rose petals. Jacques wants his potpourri to have a
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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.
Sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3