1)
y=x²+2x
we have to calculate the first derived
y´=2x+2
we equal to "0" the first derived and find out the value of "X"
2x+2=0
x=-2/2=-1
we have to calculate de second derived
y´´=2>0 ⇒we have a minimun at x=-1
we calculate y
y=(-1)²+2(-1)=1-2=-1
Answer: we have a minimum at (-1,-1)
2)
y=4x³+3x²+2
we have to calculate the first derived
y´=12x²+6x
we equalize to "0" the first derived and find out the value of "X"
12x²+6x=0
6x(2x+1)=0
6x=0 ⇒x=0
2x+1=0 ⇒x=-1/2
we have to calculate de second derived
y´´=24x+6
y``(0)=24(0)+6=6 >0 ⇒we have a minimun at x=0
y``(-1/2)=24(-1/2)+6=-12+6=-6 ⇒we have a maximum at x=-1/2
we calculate y
if x=0, y=2
if x=-1/2; y=4(-1/2)³+3(-1/2)²+2=9/4.
we have to equalize the second derived to "0" and find out the value of "x"
24x+6=0
x=-6/24=-1/4; in x=-1/4 we have an inflection point.
y=4(-1/4)³+3(-1/4)²+2=31/16
Answer: we have a minimum at (0,2), a maximum at (-1/2, 9/4) and a inflection point at (-1/4, 31/16).
3.
y=2x+3+8/x
y=(2x²+3x+8)/ x
we have to calculate the first derived
y´= [(4x+3)x-(2x²+3x+8)] / x²=(4x²+3x-2x²-3x-8) / x²=(2x²-8)/x²
we equal to "0" the first derived and find out the value of "X"
(2x²-8) / x²=0
2x²-8=0
x=⁺₋2
we have to calculate de second derived
y´´=[(4x)x²-2x(2x²-8)] / x⁴=(4x³-4x³+16x)/x⁴)=16/x³
y``(-2)=16/(-2)³=-2>0 ⇒ at x=-2 exist a maximum
y´´(2)=16/(2)³=2<0 ⇒ at x=2 exist a minimum
we calculate y
y(-2)=-4+3-4=-5
y(2)=4+3+4=11
Answer: we have a maximum at (-2,-5 ) and a minimun at (2,11)
4)
y=x³-9x²-21x+11
we have to calculate the first derived
y´=3x²-18x-21
we equal to "0" the first derived and find out the value of "X"
3x²-18x-21=0
x²-6x-7=0
x=[6⁺₋√(36+28)]/2=(6⁺₋8)/2
x₁=-1
x₂=7
we have to calculate de second derived
y´´=6x-18
y``(-1)=-6-18=-24<0 ⇒at x=-1 exist a maximum
y´´(7)=42-18=24>0 ⇒ at x=7 exist a minimum.
we calculate y
y(-1)=-1-9+21+11=22
y(7)=343-441-147+11=-234
We equalize the second derive to 0, and find out the value of "x"
6x-18=0
x=3 in x=3 exist an inflection point
y=27-81-63+11=-106
Answer: we have a minimum at (7,-234), a maximum at (-1,22) and an inflection point at (3,-106).
5)
y=9x²/³-2x+5
we have to calculate the first derived
y´=6x⁻¹/³-2
we equal to "0" the first derived and find out the value of "X"
6x⁻¹/³-2=0
x⁻¹/³=1/3
x¹/³=3
x=27
we have to calculate de second derived
y´´=-2x^(-4/3)
y´´(27)=-0.024<0 ⇒ at x=27 exist a maximum
we calculate y
y=32
Answer: we have a maximum at (27,32)
