Question

Find the point(s), if any, at which the graph of \(f\) has a horizontal tangent. $$ f(x)=\frac{x^{2}}{x^{2}+1} $$

Short answer

The graph of \(f(x)\) has a horizontal tangent at the point (0,0)

Step-by-step solution

Finding the Derivative

Find the derivative of \(f(x)\). You can use the quotient rule which states that the derivative of \(y = \frac{u}{v}\) is \(y' = \frac{(vu' - uv')}{v^2}\). The derivative of \(f(x)\) becomes \(f'(x) = \frac{(2x(x^2+1)-2x^3)}{(x^2+1)^2}\)

Reducing the Derivative

Simplify the derivative to make it easier to find the roots. In this case, it reduces to \(f'(x) = \frac{2x}{(x^2+1)^2}\)

Finding the Roots of The Derivative

Set the derivative to zero and solve for \(x\). In this case, there is one root and it is \(x = 0\)

Finding the Corresponding y-coordinate

Substitute \(x = 0\) into the original function \(f(x)\) = \(\frac{x^{2}}{x^{2}+1}\). This gives \(y = 0\) as the y-coordinate

Key concepts

Derivative Quotient Rule

When tasked with finding the slope of a curve represented by a function that is the ratio of two other functions, the derivative quotient rule is indispensable. This rule is stated as follows: for a function given by the equation \( y = \frac{u}{v} \) where both \( u \) and \( v \) are differentiable functions of \( x \) the derivative of \( y \) with respect to \( x \) is \( y' = \frac{vu' - uv'}{v^{2}} \). In our exercise, we have \( f(x) = \frac{x^{2}}{x^{2}+1} \), where \( u = x^{2} \) and \( v = x^{2}+1 \). Applying the quotient rule, the derivative simplifies to \( f'(x) = \frac{2x}{(x^{2}+1)^2} \), which is essential in finding the points where the tangent line is horizontal. It's vital to grasp the quotient rule as it allows us to handle complex rational functions adeptly and paves the way for finding critical points in calculus.

Simplifying Derivatives

Encountering derivatives that look complex at first can often be intimidating. Simplifying derivatives makes the subsequent calculus work more manageable. In the given exercise, we start with a complex-looking derivative \( f'(x) = \frac{(2x(x^2+1)-2x^3)}{(x^2+1)^2} \) and simplify by combining like terms and canceling out where possible. This step is paramount as it results in \( f'(x) = \frac{2x}{(x^2+1)^2} \) which is much easier to work with. Simplifying derivatives is not just about making the problem easier to understand; it's also about reducing the risk of computational error in the later steps of problem-solving. The ability to simplify complex expressions efficiently is a skill that students should develop to enhance their calculus toolkit.

Finding Roots of Derivative

To determine where the graph of a function has horizontal tangents, we must find the roots of its derivative. The roots are the values of \( x \) which make the derivative equal to zero, indicating points where the slope of the tangent to the graph is horizontal. In our exercise, setting the simplified derivative \( f'(x) = \frac{2x}{(x^2+1)^2} = 0 \) shows that \( x = 0 \) is the only root. This step is central in calculus when finding local maxima, minima, or inflection points. Recognizing how to locate the roots of a derivative rapidly guides students in sketching the behavior of function curves and understanding the nature of functions at specific points.

Tangent Line Calculus

In calculus, the concept of a tangent line to a curve at a given point is a linchpin. Tangent lines represent the instantaneous rate of change of the function at that very point. For a horizontal tangent, which is the focus of our exercise, the slope must be zero. Once the roots of the derivative are found, as done in our example with the root \( x = 0 \), we then find the corresponding \( y \) value by inserting this \( x \) back into the original function. This gives us the exact point on the graph where the tangent line is horizontal. In our case, when \( x = 0 \) the corresponding \( y = 0 \) as well, contributing to the point \( (0,0) \) where the graph of \( f(x) \) has a horizontal tangent. Understanding the relationship between a function's derivative and tangent lines is essential for problem-solving in differential calculus and for interpreting the graph's behavior.