Question

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

Answer 1

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))