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Determine whether the statement is true or false. Explain your answer. Find an equation of the tangent line to the graph of \(y=f(x)\) at \(x=2\) if \(f(2)=-2\) and \(f^{\prime}(2)=-1\)

Short Answer

Expert verified
True; the tangent line is \\(y = -x\\).

Step by step solution

01

Understand the Information Given

We know the point on the graph where the tangent touches it, \(x = 2\), with \(f(2) = -2\), so the point is (2, -2). We are also given the slope of the tangent line at this point, \(f'(2) = -1\).
02

Recall the General Formula for a Line

The formula for the equation of a line is \(y = mx + b\), where \m\ is the slope and \b\ is the y-intercept.
03

Plug in the Slope

Since the slope of the tangent line is given by \(f'(2) = -1\), substitute \-1\ for \(m\) in the line equation: \(y = -1x + b\).
04

Use the Point to Find the Y-Intercept

Substitute the known point (2, -2) into the equation \(y = -1x + b\) to find \(b\). This gives us \(-2 = -1(2) + b\). Solve it to get \(-2 = -2 + b\), so \(b = 0\).
05

Write the Equation of the Tangent Line

With \(m = -1\) and \(b = 0\), the equation of the tangent line is \(y = -1x \), or simply \(y = -x\).
06

Verify and State the Truth of the Statement

Since we have been able to derive the equation of the tangent line given the point and slope, the original statement is true, and the equation of the tangent line at \(x = 2\) is \(y = -x\).

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Key Concepts

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

Differential Calculus
Differential calculus is a fundamental branch of mathematics that focuses on the concept of change. It studies how functions change when their inputs change. One of the main tools in differential calculus is the derivative. This measures how much a function's output changes as you minutely change the input.
This concept is crucial because it helps in understanding the behavior of functions. This includes how they rise and fall over their domain. At its core, differential calculus gives us methods to calculate the rate at which quantities change and can be applied to various fields like physics and economics.
When dealing with curves or graphs of functions, differential calculus is used to find slopes, rates of change, and problems involving optimization. In practical situations, you can use it to determine the speed of a moving object or how quickly a disease spreads. The ability to calculate the slope at any point on a function using its derivative is a vital skill facilitated by differential calculus.
Derivative Calculation
Calculating derivatives is one of the primary tasks in differential calculus. The derivative of a function at a given point tells you the slope of the tangent line to the graph of the function at that point. This slope represents the rate of change of the function with respect to its variable.
To find a derivative, you often take the limit as the change in the variable approaches zero. For example, for a function \(f(x)\), the derivative \(f'(x)\) is defined mathematically as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
This expression gives us the instantaneous rate of change of the function. For the original exercise, finding \(f'(2)\) meant understanding how the function changes specifically when \(x=2\). When the derivative \(f'(x)\) is known, you can easily use it to find the tangent line to the function at the point \(x\).
Derivatives can be applied to solve various real-world problems, such as predicting future stock prices or understanding how populations grow over time. Understanding how to calculate and use them is a fundamental part of learning calculus.
Slope of Tangent Line
The slope of the tangent line to a point on a function is a central concept in differential calculus. It describes how steep the graph is at that particular point. The tangent line is the best linear approximation of the function near that point.
In the context of the original exercise, the slope of the tangent line, denoted as \(f'(2) = -1\), indicates that at \(x = 2\), the function is decreasing at a rate of 1 unit per unit increase in \(x\). This means if you slightly move to the right on the graph, the function goes down.
To find the equation of the tangent line, you use the point-slope formula \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. By rearranging this formula, you get the familiar linear equation \(y = mx + b\).
Understanding the slope helps to visualize how a function behaves locally. It is useful in physics for understanding velocity and acceleration, and in economics for determining marginal costs and revenues. Being able to calculate and interpret the slope of the tangent line is fundamental to mastering differential calculus.

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Most popular questions from this chapter

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0}\). You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8 ). Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0}\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow \infty} f^{\prime}(x)\) Let $$ f(x)=\left\\{\begin{array}{ll} x^{2}-16 x, & x<9 \\ \sqrt{x}, & x \geq 9 \end{array}\right. $$ Is \(f\) continuous at \(x=9 ?\) Determine whether \(f\) is differentiable at \(x=9 .\) If so, find the value of the derivative there.

Let $$ f(x)=\left\\{\begin{array}{ll} x^{2} \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{array}\right. $$ (a) Show that \(f\) is continuous at \(x=0\) (b) Use Definition 2.2 .1 to find \(f^{\prime}(0)\) (c) Find \(f^{\prime}(x)\) for \(x \neq 0\) (d) Show that \(f^{\prime}\) is not continuous at \(x=0\)

A robot moves in the positive direction along a straight line so that after \(t\) minutes its distance is \(s=6 t^{4}\) feet from the origin. (a) Find the average velocity of the robot over the interval [2,4]. (b) Find the instantaneous velocity at \(t=2.\)

Show that the segment cut off by the coordinate axes from any tangent line to the graph of \(y=1 / x\) is bisected by the point of tangency.

Recall that a function \(f\) is even if \(f(-x)=f(x)\) and odd if \(f(-x)=-f(x),\) for all \(x\) in the domain of \(f .\) Assuming that \(f\) is differentiable, prove: (a) \(f^{\prime}\) is odd if \(f\) is even (b) \(f^{\prime}\) is even if \(f\) is odd.

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