SOLUTION: Find the coordinates of the turning points of y = x^2/(x+1) . Determine in each case whether the point is a maximum or a minimum.

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Question 1192429: Find the coordinates of the turning points of y = x^2/(x+1) . Determine in each case whether the point is a maximum or a minimum.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Extreme Points of y+=+x%5E2%2F%28x%2B1%29:
If f+'%28x+%29%3E0 to the left of x=c and f'%28x+%29%3C0 to the right of x=c then x=c is a local maximum.
If f '%28x+%29%3C0 to the left of x=c and f'%28x+%29%3E0 to the right of x=c then x=c is a local minimum.

find f '%28x+%29

y+=+x%5E2%2F%28x%2B1%29..........y=f%28x%29
f%28x%29=+x%5E2%2F%28x%2B1%29
Apply the Quotient Rule: %28f%2Fg%29'=(f'%28g%29-f(g'))/g%5E2

f'%28x%29+=%282x%28x%2B1%29-x%5E2%2A1%29%2F%28x%2B1%29%5E2
f'%28x%29+=%282x%5E2%2B2x-x%5E2%29%2F%28x%2B1%29%5E2
f'%28x%29+=%28x%5E2%2B2x%29%2F%28x%2B1%29%5E2
f'%28x%29+=+%28x+%28x+%2B+2%29%29%2F%28x+%2B+1%29%5E2

f'%28x%29+=0

%28x+%28x+%2B+2%29%29%2F%28x+%2B+1%29%5E2=0 will be only if numerator equal to zero, so
x+%28x+%2B+2%29=0
=>x=0+or x=-2

then
f%280%29=+0%5E2%2F%280%2B1%29=0-> point is (0,0)
f%28-2%29=+%28-2%29%5E2%2F%28-2%2B1%29=4%2F-1=-4-> point is (-2,-4)


so we have:
Maximum: at (-2,-4)
Minimum: at (0,0)