Find the points on the graph of f where the tangent line is horizontal.

original: x/x^2+25)

After I did the quotient rule I got:

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

Then of course I know you set that equal to 0, but I do not know how to set it up or "finalize" the problem because of the denominator. Please help!

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

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

As long as x^2+25 <> 0 then you can multiply both sides by the denominator

(25-x^2) = 0

Thus
x^2 = 25
x = ±5

( Check: 25+25<>0 .. okay)

Find y when x=-5 and x=+5. These are your points

does that go into the derivative equation or the original?

To find the points on the graph of function f where the tangent line is horizontal, we need to determine when the derivative of f equals zero.

Let's start by finding the derivative of f using the quotient rule. The derivative of f(x) = x/(x^2+25) is given by:

f'(x) = [(x^2 + 25)(1) - (x)(2x)] / (x^2+25)^2

Now, to find the points where the tangent line is horizontal, we set the derivative f'(x) equal to zero and solve for x:

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

Next, simplify the equation by multiplying both sides by (x^2+25)^2:

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

Expand and collect like terms:

x^2 + 25 - 2x^2 = 0

Simplify further:

25 - x^2 = 0

Rearrange the equation to isolate x^2:

x^2 = 25

Taking the square root of both sides, we get:

x = ±√25

Therefore, the solutions for x are x = 5 and x = -5.

These are the values of x where the tangent line to the graph of f is horizontal. To find the corresponding y-values, substitute these x-values back into the original equation f(x):

For x = 5, f(5) = 5/(5^2+25) = 5/50 = 1/10.
For x = -5, f(-5) = -5/((-5)^2+25) = -5/50 = -1/10.

Hence, the points on the graph of f where the tangent line is horizontal are (5, 1/10) and (-5, -1/10).