Find the points on the graph of f where the tangent line is horizontal.
x/x^2+25
I know you have to do the quotient rule to find the derivative, but then I do not know what to do after that.
The tangent is horizontal at the point where the derivative is zero.
Solve: dy/dx = 0 when y = x/(x^2+25)
To find the points on the graph of f where the tangent line is horizontal, we need to first find the derivative of the function f and then solve for the values of x where the derivative is equal to zero.
Let's start by finding the derivative of f using the quotient rule. The quotient rule states:
d/dx(u/v) = (v * du/dx - u * dv/dx) / v^2
For the function f(x) = x/(x^2 + 25), we can rewrite it as f(x) = x * (x^2 + 25)^(-1) to apply the quotient rule.
Now, let's find the derivative f'(x):
f'(x) = [x^2 + 25 * 1 - x * 2x] / (x^2 + 25)^2
f'(x) = (x^2 + 25 - 2x^2) / (x^2 + 25)^2
f'(x) = (25 - x^2) / (x^2 + 25)^2
To find the points where the tangent line is horizontal, we need to set the derivative f'(x) equal to zero and solve for x:
(25 - x^2) / (x^2 + 25)^2 = 0
Since the numerator of the equation is zero, the expression will be zero only when the numerator is zero. That means:
25 - x^2 = 0
Solving this equation, we get:
x^2 = 25
x = ± √25
x = ± 5
So, the possible values of x where the tangent line is horizontal on the graph of f(x) = x/(x^2 + 25) are x = 5 and x = -5.
Therefore, the points on the graph of f where the tangent line is horizontal are (5, f(5)) and (-5, f(-5)). To find the corresponding y-values, substitute these x-values into the original function f(x):
When x = 5:
f(5) = 5/(5^2 + 25) = 5/50 = 1/10
When x = -5:
f(-5) = -5/((-5)^2 + 25) = -5/50 = -1/10
So, the two points on the graph of f where the tangent line is horizontal are (5, 1/10) and (-5, -1/10).