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Chapter 2: Problem 75

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

Short Answer

Expert verified
The function \(f(x) = \frac{x^{2}}{x-1}\) has a horizontal tangent line at x = 2.

Step by step solution

01

Find the Derivative of the Function

To find the derivative of the function \(f(x)=\frac{x^{2}}{x-1}\), we will need to use the quotient rule of differentiation. The rule states that if we have a function in the form of \(u(x) 梅 v(x)\), its derivative will be \(\frac{u'(x)v(x) - u(x)v'(x)}{{v(x)}^2}\). Here, \(u(x)\) is \(x^{2}\) and \(v(x)\) is \(x-1\). So, the derivative \(f'(x)\) will be: \(f'(x)=\frac{2x(x-1) - x^{2}}{(x-1)^{2}}\).
02

Set Derivative Equal to 0

We find the x-values where the derivative equals 0, as those will be the points where the tangent is horizontal. So, we need to solve the equation \(f'(x) = 0\): \(\frac{2x(x-1) - x^{2}}{(x-1)^{2}} = 0\).
03

Simplify and Solve

After simplifying the above equation, we get \(x = 2\). We got only one solution, so the function \(f(x)=\frac{x^{2}}{x-1}\) has a horizontal tangent at the point (2, \(f(2)\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
When studying calculus, one of the foundational concepts encountered is the derivative of a function. In essence, the derivative represents the rate at which a function is changing at any given point; it's the slope of the tangent line to the function's curve at a particular point. Consider riding a bicycle on a hill鈥攜our speed at any instant is like the derivative at a point on a hill-shaped graph of your distance traveled over time.

To calculate the derivative, there are various rules depending on the type of function you're dealing with. For polynomial functions (those involving powers of x), you typically use the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\). But for a ratio of two functions, as in our exercise, we must apply the quotient rule.
Quotient Rule of Differentiation
The quotient rule is a procedure for finding the derivative of a function that is formed by dividing one function by another. It is especially useful when both the numerator and the denominator of a function are also functions of x鈥攖his is frequent in real-life applications, such as in rates of change where variables are interdependent.

The quotient rule formula is \( \frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2} \) where \( u \) and \( v \) are functions of x, and \( u' \) and \( v' \) are their respective derivatives. It's crucial to accurately discern which function is the numerator (u) and which is the denominator (v) when using this rule. Misplacing these can lead to incorrect results, so careful attention is needed.

To make sure we master the rule, it helps to practice with a variety of functions, both simple and complex. This hones our ability to correctly identify the parts of the function and simplifies the process of differentiation.
Solving Equations
Once we have the derivative and we're seeking where the function has a horizontal tangent line, we encounter our next challenge鈥攕olving equations. This involves finding the value(s) of x that make the equation true. For horizontal tangents, we set the derivative equal to zero because a horizontal line has a slope of 0, and the slope is what the derivative measures at a point on a function.

There are many techniques to solve equations, including factoring, using the quadratic formula, graphing, or employing algebraic manipulation. The choice of method often depends on the form of the equation. In our example, after setting the derivative to zero, we ideally aim to simplify the equation to its most basic form, often requiring us to factor expressions or cancel common terms. This process leads us to the critical x-values where the function's rate of change鈥攁nd thus its slope鈥攊s zero. These x-values, when substituted back into the original function, give us the precise points on the graph where the tangent line is horizontal.

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Most popular questions from this chapter

Electrical Circuit The voltage \(V\) in volts of an electrical circuit is \(V=I R,\) where \(R\) is the resistance in ohms and I is the current in amperes. \(R\) is increasing at a rate of 2 ohms per second, and \(V\) is increasing at a rate of 3 volts per second. At what rate is \(I\) changing when \(V=12\) volts and \(R=4\) ohms?

Modeling Data The normal daily maximum temperatures \(T\) (in degrees Fahrenheit) for Chicago, Illinois, are shown in the table. (Source: National Oceanic and Atmospheric Administration) $$\begin{array}{|c|c|c|c|c|}\hline \text { Month } & {\text { Jan }} & {\text { Feb }} & {\text { Mar }} & {\text { Apr }} \\ \hline \text { Temperature } & {31.0} & {35.3} & {46.6} & {59.0} \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|c|}\hline \text { Month } & {\text { May }} & {\text { Jun }} & {\text { Jul }} & {\text { Aug }} \\ \hline \text { Temperature } & {70.0} & {79.7} & {84.1} & {81.9} \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|c|}\hline \text { Month } & {\text { Sep }} & {\text { Oct }} & {\text { Nov }} & {\text { Dec }} \\ \hline \text { Temperature } & {74.8} & {62.3} & {48.2} & {34.8} \\ \hline\end{array}$$ (a) Use a graphing utility to plot the data and find a model for the data of the form \(T(t)=a+b \sin (c t-d)\) where \(T\) is the temperature and \(t\) is the time in months, with \(t=1\) corresponding to January. (b) Use a graphing utility to graph the model. How well does the model fit the data? (c) Find \(T^{\prime}\) and use a graphing utility to graph \(T^{\prime}\) . (d) Based on the graph of \(T^{\prime}\) , during what times does the temperature change most rapidly? Most slowly? Do your answers agree with your observations of the temperature changes? Explain.

Proof Let \((a, b)\) be an arbitrary point on the graph of \(y=1 / x, x>0 .\) Prove that the area of the triangle formed by the tangent line through \((a, b)\) and the coordinate axes is 2 .

Comparing Methods Consider the function \(r(x)=\frac{2 x-5}{(3 x+1)^{2}}\) (a) In general, how do you find the derivative of \(h(x)=\frac{f(x)}{g(x)}\) using the Product Rule, where \(g\) is a composite function? (b) Find \(r^{\prime}(x)\) using the Product Rule. (c) Find \(r^{\prime}(x)\) using the Product Rule. (d) Which method do you prefer? Explain.

$$\begin{array}{l}{\text { Using the Alternative Form of the }} \\ {\text { Derivative In Exercises } 69-76, \text { use the the }} \\ {\text { alternative form of the derivative to find the }} \\ {\text { derivative at } x=c, \text { if it exists. }}\end{array}$$ $$f(x)=x^{3}+2 x^{2}+1, c=-2$$
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