What is (f(x))/(2x)<= (x^2+1)/(2x) ?

The solution to (f(x))/(2x)<= (x^2+1)/(2x) is Parabola with (h,k)=(0,1),p= 1/4

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(f(x))/(2x)<= (x^2+1)/(2x)

\frac{f ( x )}{2 x} \leq \frac{x ^{2} + 1}{2 x}

(h, k) = (0, 1), p = \frac{1}{4}

Solution steps

\frac{y}{2 x} \leq \frac{x ^{2} + 1}{2 x}

Rewrite

\frac{y}{2 x} - \frac{x ^{2} + 1}{2 x} \leq 0

in the standard form

4 \cdot \frac{1}{4} (y - 1) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 1), p = \frac{1}{4}

(y + 4)^{2} = - 42 (x + 2)

(\frac{2 y ^{2} + y}{4}) = x

y < x^{2} - x + 4

5 x^{2} - 3 y = 4

x = y^{2} + 4 y - 5

What is (f(x))/(2x)<= (x^2+1)/(2x) ?
The solution to (f(x))/(2x)<= (x^2+1)/(2x) is Parabola with (h,k)=(0,1),p= 1/4