Find the turning point and dertemine whether they are Maxima or minima.
i.y=x2-6x+9
ii.y=x2+5xz+z2
iii.y=x2+z2+v2
i.
y'=(x2-6x+9'=2x-6
y''=(2x-6)''=2
2x-6=0
x=3
The second derivative does not contain x , so insertion gives 2. 2 is larger than 0. So there is a mimimum at 3.
"y(3)=3^2-6\\times3+9=9-18+9=0"
Minimum turning point (3;0).
ii. y(x):
"x^2+5xz=y-z^2"
"x^2+5xz+(\\frac{5z}{2})^2=(\\frac{5z}{2})^2+y-z^2"
solution set
"x1=\\frac{-5z}{2}+(y-z^2+\\frac{25z^2}{4})^{0.5}"
"x2=\\frac{-5z}{2}-(y-z^2+\\frac{25z^2}{4})^{0.5}"
y(z)
"z^2+5xz=-x^2+y"
"z^2+5xz+(\\frac{5x}{2})^2=(\\frac{5x}{2})^2+y-x^2"
"x1=\\frac{-5x}{2}+(y-x^2+\\frac{25x^2}{4})^{0.5}"
"x2=\\frac{-5x}{2}-(y-x^2+\\frac{25x^2}{4})^{0.5}"
iii.
y(x)
"z^2+v^2=y-x^2"
"z^2=y-x^2-v^2"
"z=+-(y-x^2-v^2)^{0.5}"
y(z)
"z^2+x^2=y-v^2"
"x^2=y-z^2-v^2"
"x=+-(y-z^2-v^2)^{0.5}"
y(v)
"z^2+v^2=y-x^2"
"v^2=y-z^2-x^2"
"v=+-(y-z^2-x^2)^{0.5}"
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