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Chapter 13: Problem 8

Find the slope of the tangent line to each curve when \(x\) has the given value. Do not use a calculator. $$f(x)=6 x^{2}-4 x, x=-1$$

Short Answer

Expert verified
The slope of the tangent line is \(-16\).

Step by step solution

01

Understand the Problem

We need to find the slope of the tangent line to the curve represented by the function \( f(x) = 6x^2 - 4x \) at the point where \( x = -1 \). This involves finding the derivative of the function and evaluating it at \( x = -1 \).
02

Find the Derivative

The slope of the tangent line at a given point on the curve is the value of the derivative of the function at that point. Let's find \( f'(x) \), the derivative of \( f(x) = 6x^2 - 4x \).Using the power rule, the derivative of \( f(x) \) is calculated as follows:\[ f'(x) = \frac{d}{dx}(6x^2) - \frac{d}{dx}(4x) \]Thus,\[ f'(x) = 12x - 4 \].
03

Evaluate the Derivative at \( x = -1 \)

Now that we have \( f'(x) = 12x - 4 \), we will substitute \( x = -1 \) to find the slope of the tangent line.\[ f'(-1) = 12(-1) - 4 \].Calculate \[ f'(-1) = -12 - 4 = -16 \].
04

Conclude the Solution

The slope of the tangent line to the curve \( f(x) = 6x^2 - 4x \) at \( x = -1 \) is \(-16\). This means the line is decreasing steeply at that point on the curve.

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Key Concepts

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

Slope of a Curve
When we talk about the slope of a curve at a specific point, we are essentially referring to the steepness or incline of the curve at that point. Unlike straight lines that have a constant slope, curves can have different slopes at different points. This varying slope is because the shape of the curve changes continuously.
To find the slope of a curve at a particular point, we determine the slope of the tangent line to the curve at that point. The tangent line is a straight line that just "touches" the curve at the point of interest and has the same slope as the curve there.
  • The slope of the curve at a point gives us an idea of how the curve behaves at that exact location.
  • A positive slope means the curve is rising, while a negative slope indicates it is falling.
Finding this slope involves calculus, specifically the derivative of the function representing the curve.
Derivative of a Function
The derivative of a function is a cornerstone concept in calculus. It measures how the function's output changes concerning its input. In simpler terms, it tells us how a tiny change in the input affects the function's output.
A derivative at a specific point gives us the slope of the tangent line to the curve at that point.
When we compute the derivative of a function like \( f(x) = 6x^2 - 4x \), we find an expression for \( f'(x) \) that can give the slope of the tangent for any value of \( x \).
  • The derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).
  • Finding the derivative is essential for identifying how a variable quantity changes over time.
Understanding derivatives is crucial for tasks like determining maximum and minimum values of functions, which is vital across physics, engineering, and economics.
Power Rule in Calculus
In calculus, the power rule is a basic tool used to find the derivative of functions involving exponents. It simplifies the process of differentiation considerably and is applicable to any function of the form \( x^n \), where \( n \) is a real number.
The power rule states that if you have a function \( f(x) = x^n \), its derivative \( f'(x) \) is \( nx^{n-1} \). The power gets multiplied by the coefficient, and then you reduce the power by one.
  • For example, if \( f(x) = 6x^2 \), then according to the power rule, \( f'(x) = 2 \cdot 6x^{2-1} = 12x \).
  • This method dramatically reduces the complexity of taking derivatives and is one of the first rules of differentiation that students learn.
The power rule was applied in the original exercise to find the derivative \( f'(x) = 12x - 4 \). Understanding this rule helps to tackle more complex differentiation tasks with greater ease.

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Most popular questions from this chapter

Evaluate each limit. (a) \(\lim _{x \rightarrow 0} \sqrt{4-x^{2}}\) (b) \(\lim _{x \rightarrow 3} \sqrt{4-x^{2}}\) (c) \(\lim _{x \rightarrow 2} \sqrt{4-x^{2}}\)

Solve each problem.Use a calculator to answer each of the following. (a) From a graph of \(y=x e^{-x},\) what do you think is the value of \(\lim _{x \rightarrow \infty}\left(x e^{-x}\right) ?\) Support your answer by evaluating the function for several large values of \(x\). (b) Repeat part (a), but this time use the graph of the function \(y=x^{2} e^{-x}\) (c) On the basis of your results from parts (a) and (b), what do you think is the value of \(\lim \left(x^{n} e^{-x}\right)\) for other positive integers \(n ?\)

Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 0} \cos \frac{1}{x}\)

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Evaluate each limit. $$\begin{array}{l} \text { (a) } \lim _{x \rightarrow \pi^{-}} \cot x \\ \text { (b) } \lim _{x \rightarrow \pi^{+}} \cot x \\ \text { (c) } \lim _{x \rightarrow \pi} \cot x \end{array}$$
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