Find the equation of the tangent to the curve y = (4x)/(x + 1) which is perpendicular to the line y = x

To find the equation of the tangent to the curve \( y = \frac{4x}{x + 1} \) that is perpendicular to the line \( y = x \), we can follow these steps:

### Step 1: Identify the slope of the perpendicular line
The slope of the line \( y = x \) is 1. The slope of any line perpendicular to this line would be the negative reciprocal of 1, which is \( -1 \).

### Step 2: Find the derivative of the curve
The slope of the tangent to the curve at any point is given by the derivative of \( y = \frac{4x}{x + 1} \) with respect to \( x \).

To differentiate \( y = \frac{4x}{x + 1} \), we can use the quotient rule:

\[
\text{If } y = \frac{u(x)}{v(x)}, \text{ then } \frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}
\]

For \( y = \frac{4x}{x + 1} \), let \( u(x) = 4x \) and \( v(x) = x + 1 \). 

- \( u'(x) = 4 \)
- \( v'(x) = 1 \)

Now, applying the quotient rule:

\[
\frac{dy}{dx} = \frac{(x + 1) \cdot 4 - 4x \cdot 1}{(x + 1)^2} = \frac{4x + 4 - 4x}{(x + 1)^2} = \frac{4}{(x + 1)^2}
\]

### Step 3: Set the derivative equal to \(-1\)
We want the slope of the tangent (which is the derivative) to be \( -1 \):

\[
\frac{4}{(x + 1)^2} = -1
\]

However, this equation has no real solution since \( \frac{4}{(x + 1)^2} \) is always positive, while \( -1 \) is negative. This means that there is no point on the curve where the tangent line has a slope of \( -1 \).

### Conclusion:
Since the derivative \( \frac{4}{(x + 1)^2} \) is always positive and never equals \( -1 \), **there is no point on the curve \( y = \frac{4x}{x + 1} \) where the tangent is perpendicular to the line \( y = x \)**.

Would you like to explore a different related problem or have further questions?

Here are 5 related questions:
1. What is the geometric meaning of the slope of a tangent line?
2. How can you find the equation of a tangent line given a specific point on a curve?
3. What is the significance of the quotient rule in differentiation?
4. How would the problem change if the curve were different, say \( y = \frac{4x}{x + 2} \)?
5. Can you find the points on the curve where the tangent line has a positive slope of 1?

**Tip:** When looking for perpendicular tangents, always check the sign of the slope to determine if a solution is possible.

Learn how to find the equation of the tangent to the curve y = (4x)/(x + 1) that is perpendicular to the line y = x. This problem involves calculus concepts such as derivatives and the quotient rule. Suitable for advanced high school students.

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