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Chapter 3: Problem 18

In Exercises \(17-18\) , differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function. $$ y=f(x)=\frac{8}{\sqrt{x-2}}, \quad(x, y)=(6,4) $$

Short Answer

Expert verified
The equation of the tangent line is \( y = -x + 10 \).

Step by step solution

01

Differentiate the Function

To differentiate \( y=f(x)=\frac{8}{\sqrt{x-2}} \), first rewrite the function as \( y = 8(x-2)^{-1/2} \). Apply the power rule where the derivative \( y' \) becomes:\[ y' = 8 \cdot \left(-\frac{1}{2}\right)(x-2)^{-3/2} \cdot 1 = -4(x-2)^{-3/2} \].This simplifies to \( y' = -\frac{4}{(x-2)^{3/2}} \).
02

Evaluate the Derivative at the Point

Substitute \( x=6 \) into the derivative to find the slope of the tangent line at that point:\[ y'(6) = -\frac{4}{(6-2)^{3/2}} = -\frac{4}{4} = -1 \].Therefore, the slope of the tangent line at the point \((6,4)\) is \(-1\).
03

Use Point-Slope Form to Find the Tangent Line

With the slope \( m = -1 \) and the point \((6,4)\), use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), to find the tangent line:\[ y - 4 = -1(x - 6) \].Simplify this to get the equation of the tangent line:\[ y = -x + 10 \].

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Key Concepts

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

Tangent Line Equation
The tangent line to a function at a given point is a straight line that just "touches" the curve at that particular point. This line represents the best linear approximation of the function near that point.
The tangent line is useful in understanding the behavior of functions near a given point, especially in calculus. It essentially matches the slope of the curve at its "touching" point, showing how the function is changing there.
In this exercise, the tangent line to the function at \(x = 6\) had a calculated slope, allowing us to write an equation describing this line: \(y = -x + 10\). This equation captures the linear nature of the tangent line at point \(6,4\).
Power Rule
The power rule is a basic but powerful tool in calculus for finding derivatives. It states that the derivative of \(x^n\) is \(nx^{n-1}\), where \(n\) is any real number.
In our exercise, the function \(y=f(x)=\frac{8}{\sqrt{x-2}}\) was first rewritten using exponents: \(y = 8(x-2)^{-1/2}\). This new format allowed the straightforward application of the power rule. By applying it, we find the derivative \(y' = -\frac{4}{(x-2)^{3/2}}\).
Understanding and applying the power rule makes differentiation tasks quicker and more efficient, especially when dealing with polynomial or polynomial-like functions.
Derivative Evaluation
Once you have the derivative function, the next step is to evaluate it at a specific point. The purpose of derivative evaluation is to find the slope of the tangent line to the curve at a given point. This slope tells us how steep the curve is at that point.
In our example, after differentiating the function, we found its derivative was \(y' = -\frac{4}{(x-2)^{3/2}}\). By substituting \(x = 6\) into this derivative, the slope calculated was \(-1\). This slope is crucial to forming the tangent line equation.
Derivative evaluation is essential in calculus as it helps identify points of increasing or decreasing trends and provides insights into the curvature of graphs.
Point-Slope Form
The point-slope form is a highly effective way to write the equation of a line that has a known slope and passes through a specific point. Its format is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.
In this problem, we determined the slope \(m = -1\) and a point \((6, 4)\). By plugging these values into the point-slope form, we arrived at: \(y - 4 = -1(x - 6)\).
Upon simplifying, the resulting equation was \(y = -x + 10\).
  • Using point-slope form is advantageous as it directly incorporates both the slope and point information.
  • This makes it straightforward to derive the tangent line equation, enhancing clarity when interpreting or constructing linear graphs.

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Most popular questions from this chapter

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the volume \(V=(4 / 3) \pi r^{3}\) of a sphere when the radius changes from \(r_{0}\) to \(r_{0}+d r\)

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2} $$

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the lateral surface area \(S=\pi r \sqrt{r^{2}+h^{2}}\) of a right circular cone when the radius changes from \(r_{0}\) to \(r_{0}+d r\) and the height does not change

a. Given that \(x^{4}+4 y^{2}=1\) , find \(d y / d x\) two ways: \((1)\) by solving for \(y\) and differentiating the resulting functions in the usual way and \((2)\) by implicit differentiation. Do you get the same result each way? b. Solve the equation \(x^{4}+4 y^{2}=1\) for \(y\) and graph the resulting functions together to produce a complete graph of the equation \(x^{4}+4 y^{2}=1 .\) Then add the graphs of the first derivatives of these functions to your display. Could you have predicted the general behavior of the derivative graphs from looking at the graph of \(x^{4}+4 y^{2}=1 ?\) Could you have predicted the general behavior of the graph of \(x^{4}+4 y^{2}=1\) by looking at the derivative graphs? Give reasons for your answers.

In Exercises \(31-36,\) each function \(f(x)\) changes value when \(x\) changes from \(x_{0}\) to \(x_{0}+d x .\) Find a. the change \(\Delta f=f\left(x_{0}+d x\right)-f\left(x_{0}\right)\) b. the value of the estimate \(d f=f^{\prime}\left(x_{0}\right) d x ;\) and c. the approximation error \(|\Delta f-d f|\) $$ f(x)=x^{2}+2 x, \quad x_{0}=1, \quad d x=0.1 $$
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