4) The graph of y = (x)/(1+25x^2) for x is greater than or equal to 0 has a horizontal tangent at x = what.

So what I did was find the first derivative. I did the quotient rule and ended up with
x50x - 1+25x^2
---------------
(1+25x^2)^2

I equated that with 0 but I don't know how to solve for x without using a calculator. Someone please help.

To solve for x without using a calculator, let's simplify the numerator of the first derivative.

The numerator is: x(50x) - 1 + 25x^2
Simplifying that, we get: 50x^2 - 1 + 25x^2
Combining like terms, we have: 75x^2 - 1

Now, let's set the numerator equal to 0 and solve for x.

75x^2 - 1 = 0

To solve this quadratic equation, we can add 1 to both sides of the equation:

75x^2 = 1

Next, divide both sides of the equation by 75:

x^2 = 1/75

Finally, take the square root of both sides of the equation:

x = ±√(1/75)

Therefore, the graph of y = (x)/(1+25x^2) has horizontal tangents at x = √(1/75) and x = -√(1/75).

To find the x-value at which the graph of the function y = x/(1+25x^2) has a horizontal tangent, you need to solve the equation obtained by setting the derivative equal to zero. Let's simplify the derivative equation you obtained:

dy/dx = (x(50x) - (1+25x^2))/((1+25x^2)^2)
dy/dx = (50x^2 - 1 - 25x^2)/((1+25x^2)^2)
dy/dx = (25x^2 - 1)/((1+25x^2)^2)

Now, set this equation equal to zero and solve for x:

(25x^2 - 1)/((1+25x^2)^2) = 0

To find the values of x that satisfy this equation, we can first set the numerator equal to zero:

25x^2 - 1 = 0

Solving this quadratic equation, we have:

x^2 = 1/25
x = ±sqrt(1/25)
x = ±1/5

Now, we need to check which of these values make the entire denominator (1+25x^2)^2 non-zero. Both values of x = ±1/5 satisfy this condition.

Therefore, the graph of y = x/(1+25x^2) has a horizontal tangent at x = 1/5 and x = -1/5.