Find the point(s) on the graph of f where the tangent line is horizontal. (If an answer does not exist, enter DNE.)
f(x) = 
x/x2+16
(x, y)	 = 	(	) (smaller x-value)
(x, y)	 = 	(	) (larger x-value)


To find the point(s) where the tangent line to the graph of the function \( f(x) = \frac{x}{x^2 + 16} \) is horizontal, we need to:

1. Find the derivative of \( f(x) \) to determine the slope of the tangent line.
2. Set the derivative equal to 0 and solve for \( x \), since a horizontal tangent line has a slope of 0.
3. Substitute the \( x \)-values back into the original function to find the corresponding \( y \)-values.

### Step 1: Find the derivative of \( f(x) \)

We'll use the quotient rule, which states:
\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
\]
where \( u = x \) and \( v = x^2 + 16 \).

- \( u' = 1 \)
- \( v' = 2x \)

Thus, the derivative \( f'(x) \) is:
\[
f'(x) = \frac{(1)(x^2 + 16) - (x)(2x)}{(x^2 + 16)^2} = \frac{x^2 + 16 - 2x^2}{(x^2 + 16)^2} = \frac{-x^2 + 16}{(x^2 + 16)^2}
\]

### Step 2: Set the derivative equal to 0

For the tangent line to be horizontal, we need \( f'(x) = 0 \), so:
\[
\frac{-x^2 + 16}{(x^2 + 16)^2} = 0
\]

This equation is equal to zero when the numerator is zero:
\[
-x^2 + 16 = 0
\]
\[
x^2 = 16
\]
\[
x = \pm 4
\]

### Step 3: Find the corresponding \( y \)-values

Now substitute \( x = 4 \) and \( x = -4 \) into the original function to find the \( y \)-values.

For \( x = 4 \):
\[
f(4) = \frac{4}{4^2 + 16} = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8}
\]

For \( x = -4 \):
\[
f(-4) = \frac{-4}{(-4)^2 + 16} = \frac{-4}{16 + 16} = \frac{-4}{32} = \frac{-1}{8}
\]

### Final Answer:

- The point with the smaller \( x \)-value is \( (-4, -\frac{1}{8}) \).
- The point with the larger \( x \)-value is \( (4, \frac{1}{8}) \).

Thus, the points where the tangent line is horizontal are:
\[
(x, y) = (-4, -\frac{1}{8}), \quad (x, y) = (4, \frac{1}{8})
\]

Would you like further details or have any questions?

### Related Questions:
1. What is the second derivative of \( f(x) = \frac{x}{x^2 + 16} \)?
2. How do we determine if the points found are local maxima or minima?
3. Can you find the concavity of the function \( f(x) \)?
4. What is the asymptotic behavior of \( f(x) \) as \( x \to \infty \)?
5. How do horizontal tangents relate to critical points of a function?

**Tip:** The quotient rule is essential for finding derivatives of functions that are ratios of two other functions.

Find the point(s) on the graph of f where the tangent line is horizontal. (If an answer does not exist, enter DNE.) f(x) = x/(x^2+16) (x, y) = ( ) (smaller x-value) (x, y) = ( ) (larger x-value)

This problem involves finding the points where the tangent line to the graph of f(x) = x/(x^2 + 16) is horizontal. It covers key calculus concepts like the quotient rule and derivative analysis. Learn how to solve for the x-values and corresponding y-values, suitable for undergraduate calculus students.

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