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Chapter 3: Problem 13

In Exercises \(13-16,\) differentiate the functions and find the slope of the tangent line at the given value of the independent variable. $$ f(x)=x+\frac{9}{x}, \quad x=-3 $$

Short Answer

Expert verified
The slope of the tangent line at \( x = -3 \) is 0.

Step by step solution

01

Differentiate the Function

The function is given as \( f(x) = x + \frac{9}{x} \). To find the derivative, apply the power rule and the derivative of a fraction rule. The derivative of \( x \) is 1, and the derivative of \( \frac{9}{x} \) can be written as \( 9x^{-1} \), which differentiates to \(-9x^{-2} = -\frac{9}{x^2}\). Thus, \( f'(x) = 1 - \frac{9}{x^2} \).
02

Evaluate the Derivative at the Given Value

To find the slope of the tangent line at \( x = -3 \), substitute \( -3 \) into the derivative: \( f'(-3) = 1 - \frac{9}{(-3)^2} \). Calculating \(-3^2 = 9\), we have \( f'(-3) = 1 - \frac{9}{9} = 1 - 1 = 0 \).
03

Interpret the Result

The value of \( f'(-3) = 0 \) indicates that the slope of the tangent line to the curve at \( x = -3 \) is 0. This means that the tangent line is horizontal at this point.

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Key Concepts

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

Understanding Tangent Lines
A tangent line is a straight line that just "touches" a curve at a particular point, without crossing it. It essentially "hugs" the curve at that point, representing the best linear approximation to the curve around that particular point. The slope of the tangent line is crucial because it indicates how steep the curve is at that specific location.

When you find the derivative of a function at a certain point, you're determining the slope of the tangent line to the function at that point. A tangent line's slope can tell you if the curve is rising, falling, or staying flat. If the slope is 0, like in our exercise, the tangent line is horizontal. This implies that the curve has a "flat spot" at that point, which is neither rising nor falling. Understanding the behavior of tangent lines helps in sketching charts and analyzing functions.
Applying the Power Rule for Derivatives
The power rule is one of the fundamental techniques in calculus for finding derivatives. It's exceptionally user-friendly and plays a vital role in differentiating polynomial functions. The rule states that if you have a term of the form \(x^n\), its derivative will be \(nx^{n-1}\).

Let's apply the power rule to differentiate a simple function. Suppose you have a function \(f(x) = x\), which can be thought of as \(x^1\). Using the power rule, its derivative is \(1\times x^{0}=1\). In our exercise, the term \(\frac{9}{x}\) was rewritten as \(9x^{-1}\) to make it compatible with the power rule. Its derivative was then straightforward to compute as \(-9x^{-2}\). This shows the importance of rewriting terms with negative exponents when using the power rule.
Finding the Derivative of a Fraction
When differentiating functions involving fractions, it often helps to rewrite them. For example, the term \(\frac{9}{x}\) can be expressed as \(9x^{-1}\). This transformation allows for easier application of the power rule. Calculus is all about making expressions manageable, enabling us to apply differentiation rules correctly.

Finding the derivative of \(\frac{9}{x}\) involves recognizing the power of \(x\) as \(-1\). Using the power rule here, its derivative becomes \(-9x^{-2}\). This translates back to \(-\frac{9}{x^2}\) in fraction form. Transforming problems into a recognizable pattern allows for effective use of derivative rules, making the problem-solving process smoother and more intuitive.

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Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ y^{3}+\cos x y=x^{2}, \quad P(1,0) $$

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 \(|\Delta f-d f|\) $$ f(x)=x^{3}-2 x+3, \quad x_{0}=2, \quad d x=0.1 $$

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=2 t^{2}+3, \quad y=t^{4}, \quad t=-1 $$

In Exercises \(81-86,\) find a parametrization for the curve. the ray (half line) with initial point \((2,3)\) that passes through the point \((-1,-1)\)

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 \(|\Delta f-d f|\) $$ f(x)=x^{3}-x, \quad x_{0}=1, \quad d x=0.1 $$
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