Find the coordinates of the point on the curve y = x / (1 + x²) where the tangent to the curve has t

Q: Wait, why are we differentiating the derivative? Isn't that just the second derivative? Does that help us find the maximum slope?

Exactly! You have hit the nail on the head. To maximize the slope, we treat the slope function $m(x)$ as a function itself. Finding the derivative of the slope (which is the second derivative of the original curve, $y''$) and setting it to zero is the standard optimization procedure to find the local maxima and minima of the slope.

Q: Why do we get three critical points ($x = 0$ and $x = \pm\sqrt{3}$)? What do they physically represent?

Great observation. The points $x = \pm\sqrt{3}$ are where the slope is at a local minimum (the steepest downward slope), while $x = 0$ is where the slope is at a local maximum (the steepest upward slope). The curve is symmetric, so it makes sense to have these distinct points of extreme steepness.

Q: Is it possible for the slope to be negative? I thought we were looking for the 'greatest' slope.

Yes, the slope function $m(x) = \frac{1-x^2}{(1+x^2)^2}$ can definitely be negative. That just means the curve is going downhill at those points. When we say 'greatest' slope, we mean the algebraic maximum. Since $1 > -1/8$, the value $1$ at $x=0$ is indeed the maximum.

Question

Find the coordinates of the point on the curve $y = \frac{x}{1 + x^2}$ where the tangent to the curve has the greatest slope.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The path is defined by the function:

$$y = \frac{x}{1 + x^2}$$

Our objective is to identify the point on this curve where the incline, represented by the slope $m = \frac{dy}{dx}$, reaches its absolute maximum value.

### Defining the Slope Function

To determine the steepness, we calculate the first derivative of the function using the quotient rule, where $u = x$ and $v = 1 + x^2$. The quotient rule formula is $\frac{v u' - u v'}{v^2}$.

Substituting our terms, we obtain:

$$m = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2}$$

Simplifying this expression yields the slope function:

$$m(x) = \frac{1 - x^2}{(1 + x^2)^2}$$

### The Optimization Challenge

To maximize the slope $m(x)$, we must find its derivative with respect to $x$ and set it to zero. This requires calculating the second derivative of the original function:

$$\frac{dm}{dx} = \frac{d}{dx} \left( \frac{1 - x^2}{(1 + x^2)^2} \right)$$

Applying the quotient rule once more, we get:

$$\frac{dm}{dx} = \frac{-2x(1 + x^2)^2 - (1 - x^2) \cdot 2(1 + x^2) \cdot 2x}{(1 + x^2)^4}$$

### The Algebra Gauntlet

We simplify the expression by factoring out the common term $2x(1 + x^2)$ from the numerator:

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

Simplifying the bracketed term $-(1 + x^2) - 2 + 2x^2$ results in $x^2 - 3$. After canceling the common factor $(1 + x^2)$, we arrive at the elegant derivative:

$$\frac{dm}{dx} = \frac{2x(x^2 - 3)}{(1 + x^2)^3}$$

### Identifying the Winner

To find the critical points, we set $\frac{dm}{dx} = 0$. Since the denominator is always positive, we solve $2x(x^2 - 3) = 0$, yielding critical points at $x = 0$, $x = \sqrt{3}$, and $x = -\sqrt{3}$.

We now evaluate the slope $m(x)$ at these points:

1. At $x = 0$, $m = \frac{1 - 0}{(1 + 0)^2} = 1$.
2. At $x = \pm\sqrt{3}$, $m = \frac{1 - 3}{(1 + 3)^2} = \frac{-2}{16} = -\frac{1}{8}$.

Comparing these values, the maximum slope is $1$ at $x = 0$. Substituting $x = 0$ into the original function $y = \frac{x}{1 + x^2}$, we find $y = 0$.

The point of greatest slope is the origin, **$(0, 0)$**.

Animated Solution for Mathematics - Differentiation: Find the coordinates of the point on the curve $y = \frac{x}{1 + x ^{2}}$ where the tangent to the curve has the greatest slope.

Visualized Solution

Visualizing the Curve $y = \frac{x}{1 + x ^{2}}$

Finding the Slope Function $m (x)$

Calculating the First Derivative

Maximizing the Slope

Differentiating the Slope $m (x)$

Simplifying the Derivative

Final Form of the Second Derivative

Finding Critical Points

Evaluating the Slopes

Identifying the Maximum Slope

The Final Coordinates

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Defining the Slope Function

The Optimization Challenge

The Algebra Gauntlet

Identifying the Winner

Similar Questions

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point:

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Find the shortest distance of the point $(0, c)$ from the parabola $y = x^{2}$ where $0 \leq c \leq 5$ .

The maximum area of a triangle whose one vertex is at $(0, 0)$ and the other two vertices lie on the curve $y = - 2 x^{2} + 54$ at points $(x, y)$ and $(- x, y)$ where $y > 0$ is :

The local maximum value of the function $f (x) = (\frac{2}{x})^{x^{2}}, x > 0$ , is

Determine the points of maxima and minima of the function $f (x) = \frac{1}{8} ln x - b x + x^{2}, x > 0$ , where $b \geq 0$ is a constant.

The function $f (x) = 2∣ x ∣ + ∣ x + 2∣ - ∣∣ x + 2∣ - 2∣ x ∣∣$ has a local minimum or a local maximum at $x =$

Let $x$ and $y$ be two real variables such that $x > 0$ and $x y = 1$ . Find the minimum value of $x + y$ .

The number of critical points of the function $f (x) = (x - 2)^{2/3} (2 x + 1)$ is

The shortest distance between the point $(\frac{3}{2}, 0)$ and the curve $y = x, (x > 0)$ is :

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the coordinates of the point on the curve y=1+x2x​ where the tangent to the curve has the greatest slope.

Visualized Solution

Visualizing the Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=1+x2x​

Finding the Slope Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m(x)