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Chapter 3: Problem 55

In Exercises \(55-58\), plot the graph of the function. $$ f(t)=\frac{\sqrt{t^{2}+1}}{t-1} $$

Short Answer

Expert verified
The domain of the function \(f(t) = \frac{\sqrt{t^2 + 1}}{t - 1}\) is \(D = (-\infty, 1) \cup (1, \infty)\) and the range is \(R = (-\infty, \infty)\). There is a vertical asymptote at t=1, and horizontal asymptotes at f(t) = 1 and f(t) = -1. As t approaches infinity, the function approaches f(t) = 1, and as t approaches negative infinity, the function approaches f(t) = -1. There is a critical point at t=0.

Step by step solution

01

Identify the domain and range

The domain of a function is the set of possible inputs (t-values) for which the function is defined. The range is the set of possible outputs (f(t)-values) the function can have. Looking at the function \(f(t) = \frac{\sqrt{t^2 + 1}}{t - 1}\), we notice that the function will be undefined when the denominator is equal to 0. Therefore, we can't have the input t=1. The domain for this function is \(D = (-\infty, 1) \cup (1, \infty)\). To find the range, we need to see the possible outputs the function can have. Since the square root expression is always non-negative, and we are dividing by something non-zero, the function can have any real number value. The range is thus \(R = (-\infty, \infty)\).
02

Identify vertical and horizontal asymptotes

A vertical asymptote occurs when the function approaches infinity or negative infinity as the input approaches a certain value. Since the denominator becomes zero at t=1, there must be a vertical asymptote at t=1. To find horizontal asymptotes, we need to investigate the limits of the function. We consider the limits when t approaches positive and negative infinity: \(\lim_{t \to \infty} f(t) = \lim_{t \to \infty} \frac{\sqrt{t^2 + 1}}{t - 1} = 1\) \(\lim_{t \to -\infty} f(t) = \lim_{t \to -\infty} \frac{\sqrt{t^2 + 1}}{t - 1} = -1\) Thus, there are horizontal asymptotes at f(t) = 1 and f(t) = -1.
03

Determine end behavior

We've already found the horizontal asymptotes in the previous step, which describe the function's end behavior. As t approaches infinity, the function approaches 1 (f(t) = 1). As t approaches negative infinity, the function approaches -1 (f(t) = -1).
04

Identify critical points

Critical points occur where the function's derivative is zero or undefined. First, we find the derivative of f(t) using the quotient rule: \(f'(t) = \frac{(t-1)\frac{t}{\sqrt{t^2+1}} - \sqrt{t^2+1}}{(t-1)^2}\) We can now set this derivative equal to zero and solve for t to find the critical points: \(\frac{(t-1)\frac{t}{\sqrt{t^2+1}} - \sqrt{t^2+1}}{(t-1)^2} = 0\) This equation is rather complex to solve algebraically, but we can observe that there is a critical point at t=0. Now that we have found the domain, range, asymptotes, end behavior, and critical points, we provide the student with these insights so they can graph the function themselves. With practice, they'll become more comfortable graphing functions like this one.

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Key Concepts

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

Domain and Range of a Function
Understanding the domain and range of a function is critical for graphing. The domain is essentially the set of all possible 'input' values that a function can accept without causing issues, like division by zero or taking the square root of a negative number.

For the particular function given, \(f(t) = \frac{\sqrt{t^2 + 1}}{t - 1}\), we see that the domain excludes \(t=1\) because the function becomes undefined at that point. Thus, the domain is \(D = (-\infty, 1) \cup (1, \infty)\).

The range refers to the set of all 'output' or 'result' values that a function can produce. In this case, since square roots yield non-negative results and \(t-1\) cannot be zero, \(f(t)\) can take on any real number value, making the range \(R = (-\infty, \infty)\).
Vertical and Horizontal Asymptotes
Asymptotes are lines that a graph approaches but does not touch or cross. A vertical asymptote appears when the function grows without bound as the input nears a certain value.

With \(f(t)\), this happens as \(t\) approaches 1, because the denominator of the function approaches zero. This means there is a vertical asymptote located at \(t=1\).

Horizontal asymptotes occur as \(t\) goes to infinity or negative infinity and the function levels off. This function approaches 1 when \(t\) heads towards positive infinity and approaches -1 for negative infinity, leading to horizontal asymptotes at \(f(t) = 1\) and \(f(t) = -1\).
Limits and End Behavior
To grasp the end behavior of a function, one can calculate the limits as the input values \(t\) approach infinity or negative infinity. This reveals how the function behaves at the extreme ends of the domain.

For our function \(f(t)\), the limit calculations show that as \(t\) becomes very large in the positive sense, \(f(t)\) gets close to 1, and as \(t\) becomes very large in the negative sense, \(f(t)\) gets close to -1.

These limits align with the horizontal asymptotes, which delineate the function's tendency to flatten out at those values, hence summarizing its end behavior.
Critical Points and Derivative Analysis
Critical points are where a function's graph has a horizontal tangent or it's not differentiable, often corresponding to peaks, troughs, or inflection points on the graph. Derivative analysis involves finding the function's derivative and solving for points where this derivative is zero or undefined.

In our function's case, the derivative \(f'(t)\) gets complicated, but crucially, we find a critical point at \(t=0\). This spot is key for identifying notable features of the function's graph, such as relative maxima or minima and possible inflection points. These points, along with the function's asymptotes, domain, and range, provide a comprehensive framework for sketching an accurate graph of the function.

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Most popular questions from this chapter

Sketch the graph of a function having the given properties. $$ \begin{array}{l} f(2)=3, f^{\prime}(2)=0, f^{\prime}(x)<0 \text { on }(-\infty, 0) \cup(2, \infty), \\ f^{\prime}(x)>0 \text { on }(0,2), \lim _{x \rightarrow 0^{-}} f(x)=-\infty, \lim _{x \rightarrow 0^{+}} f(x)=-\infty, \\ \lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow \infty} f(x)=1, f^{\prime \prime}(x)<0 \text { on } \\ (-\infty, 0) \cup(0,3), f^{\prime \prime}(x)>0 \text { on }(3, \infty) \end{array} $$

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