Question

Horizontal Tangent Line In Exercises \(73-76\) , determine the point(s) at which the graph of the function has a horizontal tangent line. $$ f(x)=\frac{x^{2}}{x^{2}+1} $$

Short answer

The points at which the graph of function has a horizontal tangent line are \(x=0\) and \(x=\pm 1\).

Step-by-step solution

Find the derivative

We need to determine the derivative, \(f'(x)\), of the given function. Using the quotient rule for differentiation, which states that the derivative of a quotient of two functions is \( \frac{u'v-uv'}{v^2} \), where \(u(x)\) is the numerator and \(v(x)\) is the denominator, we get \(f'(x)=\frac{(2x)(x^{2}+1)-(x^{2})(2x)}{(x^{2}+1)^{2}}=\frac{2x^{2}+2x-{2x^{3}}}{(x^{2}+1)^{2}} = \frac{{2x(1-x^{2})}}{(x^{2}+1)^{2}}\).

Solve for when the derivative equals 0

To find where the slope of the tangent line is horizontal, we set \(f'(x) = 0\) and solve for \(x\). Therefore, \(2x (1 - x^{2}) = 0\). Solving this gives \(x=0\) and \(x=\pm 1\).

Verify the solutions

It's necessary to confirm that the solutions obtained correspond to horizontal tangents. Calculating the second derivative, \(f''(x)\), and evaluating it at the obtained points will confirm if the function has a maximum, minimum or inflection point at these points. If \(f''(x)\) at any of these points is non-zero, that point is a point of horizontal tangent.

Key concepts

Quotient Rule for Differentiation

When dealing with the derivative of a function presented as a ratio of two functions, the quotient rule for differentiation is an essential concept. The quotient rule is a formal rule for differentiating problems where one function is divided by another. It states that if you have a function which can be expressed as a fraction, where the numerator is a function labeled as \( u(x) \) and the denominator as \( v(x) \), the derivative \( f'(x) \) of the function \( f(x) = \frac{u(x)}{v(x)} \) is given by:

\[\begin{equation}f'(x) = \frac{u'v - uv'}{v^2}\end{equation}\]

The numerator of this quotient involves a subtraction: the derivative of the top function \( u \) times the bottom function \( v \) minus the top function \( u \) times the derivative of the bottom function \( v' \). The entire expression is then divided by the square of the bottom function \( v^2 \). This formula is crucial when calculating the derivative of a function that is not just a simple polynomial or monomial but a quotient of two functions, as seen in the example of finding a horizontal tangent line to \( f(x)=\frac{x^{2}}{x^{2}+1} \).

Finding Derivative of a Function

To find the derivative of a function means to determine the rate at which the function's value changes at any given point. Differentiation is a fundamental tool in calculus that allows us to calculate this rate of change. In the exercise provided, the derivative of the function \( f(x) \) gives us the slope of the tangent line to the curve at any point \( x \).

For the given function \( f(x)=\frac{x^{2}}{x^{2}+1} \), we applied the quotient rule for differentiation to find \( f'(x) \), which allows us to understand how the function behaves and, more specifically, where it may have horizontal tangent lines. A horizontal tangent line occurs where the slope of the function, or its derivative \( f'(x) \), is equal to zero. This is because the slope of a horizontal line in the Cartesian plane is zero, indicating no change in the y-value as x increases or decreases.

Solving for Critical Points

Critical points of a function are where the derivative of the function is either zero or undefined. These points are significant because they can indicate where the function may have potential maxima, minima, or points of inflection, which are key aspects in understanding the function’s graph.

To solve for critical points, one must first find the derivative of the function, as we did using the quotient rule. Then, set this derivative equal to zero and solve for the variable. In our exercise, setting \( f'(x) = 0 \) and solving the equation gave us the critical points at \( x = 0 \) and \( x = \pm 1 \). Not all critical points correspond to horizontal tangents – they must be verified by other means, such as using the second derivative test or by inspecting the nature of the function around those points.

Second Derivative Test

The second derivative test is used to determine the concavity of a function at a given point and to help confirm the nature of critical points found when solving for where the first derivative of a function equals zero. Concavity refers to the direction in which a function curves, either upwards like a cup (concave up) or downwards like a frown (concave down).

The sign of the second derivative at a critical point can tell us whether that point is a local maximum (concave down), local minimum (concave up), or neither (in which case the test is inconclusive). To apply this test, you take the second derivative \( f''(x) \) of the function and plug in the critical points. In our exercise, we calculated the second derivative to verify whether the points \( x = 0 \) and \( x = \pm 1 \) are indeed points of horizontal tangents by confirming they correspond to points where the function's slope changes direction, which would be indicated by a non-zero second derivative.