Where is the slope of the function f(x)=5/x^2+2 a maximum?

Answer 1

I will assume you meant
f(x) = 5/(x^2 + 2) = 5(x^2 + 2)^-1
f ' (x) = -5(x^2 + 2)^-2 (2x) <---- slope of the tangent
f '' (x) = -5(x^2 + 2)^-2 (2) + (2x)(10(x^2 + 2)^-3 (2x)
= -10(x^2 + 2)^-2 + 40x^2(x^2 + 2)^-3
= -10(x^2 + 2)^-3 [ x^2 + 2 - 4x^2] = 0 for a max/min of the slope

-3x^2 + 2 = 0
x^2 = 2/3 or 6/9
x = ± √6 / 3
y = 5/(2/3 + 2) = 15/8

So you are looking at (√6/3 , 15/8) and (-√6/3 , 15/8)

look at the graph of y = 5/(x^2 + 2)

www.wolframalpha.com/input/?i=graph+y+%3D+5%2F%28x%5E2+%2B+2%29+from+-5+to+5

If you visualize tangents drawn to the curve, they will be steepest (max slope) at (√6/3, 15/8)