Find the local max and min values of f(x)=x^2/x-1 using both first and second derivative tests

Hmm, the 2nd derivitve is good for finding concavity

let's find the max and min points
that is where the first derivitive is equal to 0
remember the difference quotient

so
f'(x)=(x^2-2x)/(x^2-2x+1)
find where it equals 0
set numerator equal to 0
0=x^2-2x
0=x(x-2)
0=x
0=x-2
2=x

so at 0 and 2 are the min and max
find if the signs go from negative to positive (min) or from positive to negative (max) at those points

f'(-1)>0
f'(1.5)<0
f'(3)>0

so at x=0, the sign go from positive to negative (local maximum)
at x=2, the sign go from negative to positive (local minimum)

we can take the 2nd derivitive to see the inflection points
f''(x)=2/((x-1)^3)
where does it equal 0?
it doesn't
so no inflection point
but, we can test it at x=0 and x=2
at x=0, we get f''(0)<0 so it is concave down. that means that x=0 being a max makes sense
at x=2, we get f''(2)>0 so it is concave up. that means that x=2 being a max make sense

local max is at x=0 (the point (0,0))
local min is at x=2 (the point (2,4))

If two sides of a triangle are congruent, then the angles opposite those sides are congruent. the converse of the base angles theorem, states that if two angles of a triangle are congruent, then sides opposite those angles are congruent.