3^2x+1=9^2x-1 solve the equation
2 answers:
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So the line that is the limit is where the equation crosses the y axis or where x=0
so
y=(1/2(0)-2)^2+5
y=(-2)^2+5
y=4+5
y=9
at y=9
the upper bound is y=9
alrighty
we will do
let's call the curve that is 6th degree f(x)
and the y=9, g(x)
f(x)=(1/2x-2)^6+5
g(x)=9
find where they intersect
9=(1/2x-2)^6+5
4=(1/2x-2)^6
fancy math
x=8
so the area under the curve will be
because g(x) is above f(x)
so
using our calculator because I can't figure out what
is
or about 21.33333333
so
y=(1/2(0)-2)^2+5
y=(-2)^2+5
y=4+5
y=9
at y=9
the upper bound is y=9
alrighty
we will do
let's call the curve that is 6th degree f(x)
and the y=9, g(x)
f(x)=(1/2x-2)^6+5
g(x)=9
find where they intersect
9=(1/2x-2)^6+5
4=(1/2x-2)^6
fancy math
x=8
so the area under the curve will be
so
using our calculator because I can't figure out what