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Chapter 2: Problem 63

Evaluate the derivative of the function at the given point. Use a graphing utility to verify your result. $$ \begin{array}{ll} \underline{\text { Function }} & \underline{\text { Point }} \\ f(t)=\frac{3 t+2}{t-1}& \quad(0,-2) \end{array} $$

Short Answer

Expert verified
The derivative of the function at the point (0,-2) is -5

Step by step solution

01

Differentiate the function

The function \(f(t)=\frac{3t+2}{t-1}\) is a quotient of two functions; hence we will apply the Quotient Rule for derivatives. The Quotient Rule states that the derivative of \(\frac{u}{v}\) is \(\frac{vu' - uv'}{v^2}\) where u and v are functions of t. Here, \(u(t) = 3t + 2\), \(v(t) = t - 1\). So, \(u'(t)=3\) and \(v'(t)=1\). Placing these values into the Quotient Rule, the derivative \(f'(t)\) is calculated as follows: \(f'(t) = \frac{(t-1)3 - (3t+2)1}{(t-1)^2}\). Simplifying we get \(f'(t) = \frac{3t-3 - 3t - 2}{{(t-1)}^2} = \frac{-5}{{(t-1)}^2}\)
02

Evaluate the derivative at the point

Now we need to evaluate \(f'(t)\) at the given point. The x-coordinate of the given point is 0. Substituting this into \(f'(t)\), we get \(f'(0) = \frac{-5}{{(0-1)}^2}\). Simplifying, we find \(f'(0) = -5\)

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Key Concepts

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

Differentiate Quotient Functions
When studying calculus, one of the essential skills is to differentiate quotient functions. These are functions where one polynomial is divided by another, often expressed as \( f(t) = \frac{u(t)}{v(t)} \). The rule that enables us to find the derivative of such a function is called the Quotient Rule.

To apply the Quotient Rule, first we identify the numerator function (\

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Most popular questions from this chapter

(a) Show that the derivative of an odd function is even. That is, if \(f(-x)=-f(x)\), then \(f^{\prime}(-x)=f^{\prime}(x)\). (b) Show that the derivative of an even function is odd. That is, if \(f(-x)=f(x)\), then \(f^{\prime}(-x)=-f^{\prime}(x)\).

Use a graphing utility to sketch the intersecting graphs of the equations and show that they are orthogonal. [Two graphs are orthogonal if at their point(s) of intersection their tangent lines are perpendicular to each other.] $$ \begin{aligned} &y^{2}=x^{3} \\ &2 x^{2}+3 y^{2}=5 \end{aligned} $$

(a) use a graphing utility to find the derivative of the function at the given point, (b) find an equation of the tangent line to the graph of the function at the given point, and (c) use the utility to graph the function and its tangent line in the same viewing window. $$ y=\left(t^{2}-9\right) \sqrt{t+2}, \quad(2,-10) $$

A buoy oscillates in simple harmonic motion \(y=A \cos \omega t\) as waves move past it. The buoy moves a total of \(3.5\) feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds. (a) Write an equation describing the motion of the buoy if it is at its high point at \(t=0\). (b) Determine the velocity of the buoy as a function of \(t\).

The cost of producing \(x\) units of a product is \(C=60 x+1350\). For one week management determined the number of units produced at the end of \(t\) hours during an eight-hour shift. The average values of \(x\) for the week are shown in the table. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{t} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \boldsymbol{x} & 0 & 16 & 60 & 130 & 205 & 271 & 336 & 384 & 392 \\ \hline \end{array} $$ (a) Use a graphing utility to fit a cubic model to the data. (b) Use the Chain Rule to find \(d C / d t\). (c) Explain why the cost function is not increasing at a constant rate during the 8 -hour shift.
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