SOLUTION: Find the minimum value of \frac{x^2}{x + 2} for x

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Question 1209911: Find the minimum value of
\frac{x^2}{x + 2}
for x > 2.
Found 3 solutions by greenestamps, math_tutor2020, mccravyedwin:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!

As x goes to positive infinity, the "+2" becomes insignificant and the expression approaches .

It should be clear that there is no maximum value of the expression.

ANSWER: No maximum value.

Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

I'm assuming you meant to say "for x > -2" since x = -2 is the vertical asymptote.
This is because the x+2 in the denominator has us go from x+2 = 0 to x = -2.
You can use a graphing tool like GeoGebra or Desmos to verify.

You can use a graphing calculator to quickly find the local min or you can use differential calculus. I'll go with the 2nd option.

f(x) = (x^2)/(x+2)
f(x) = x^2(x+2)^(-1)
f'(x) = 2x(x+2)^(-1) + x^2*(-1)*(x+2)^(-2) .... product rule
f'(x) = 2x(x+2)(x+2)^(-2) - x^2*(x+2)^(-2)
f'(x) = (2x^2+4x)(x+2)^(-2) - x^2(x+2)^(-2)
f'(x) = (2x^2+4x - x^2)(x+2)^(-2)
f'(x) = (x^2+4x)/( (x+2)^2 )

Set the derivative equal to 0 so we can determine the critical values.
f'(x) = 0
(x^2+4x)/( (x+2)^2 ) = 0
x^2+4x = 0
x(x+4) = 0
x = 0 or x+4 = 0
x = 0 or x = -4

The critical points occur when x = 0 and when x = -4.
Use either the 1st derivative test, or 2nd derivative test, to determine that a local max occurs when x = -4 and a local min occurs when x = 0.
I'll let the student handle this part.

Plug x = 0 back into the original expression.
(x^2)/(x+2) = (0^2)/(0+2) = 0
Therefore the local min on the interval x > -2 is at (0,0)
y = 0 is the smallest output on this interval.

Answer by mccravyedwin(408)   (Show Source):
You can put this solution on YOUR website!

What kind of crazy nonsense problem is this? Find the minimum value of for x > 2. That's this graph below, and we're only looking at the part where x is greater than 2. That's right of the green line at x = 2. But then it increases forever there. A minimum value???? If it were for then the minimum value would be 1 at the point (2,1). But it's x > 2, where it only gets larger and larger. Minimum value???? This is pure nuts!!! Edwin

SOLUTION: Find the minimum value of \frac{x^2}{x + 2} for x > 2.