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Chapter 5: Problem 49

In Exercises \(49-56\), find an equation of the tangent line to the graph of the function at the given point.\(f(x)=e^{1-x}, \quad(1,1)\)

Short Answer

Expert verified
The equation of the tangent line to the graph of the function at the given point is \(y = 2 - x\).

Step by step solution

01

Find the Derivative

Find the derivative of \(f(x)\) by applying the chain rule. Given that the derivative of \(e^x\) is \(e^x\) and the derivative of \(1-x\) is -1, the derivative of the function \(f'(x) = e^{1 - x} \cdot -1 = - e^{1 - x}\).
02

Evaluate the Derivative at Given Point

Find the value of the slope of the tangent line, which is the derivative of the function evaluated at the given point. Plug x = 1 into the derivative to get \(f'(1) = -e^{1 - 1} = -e^{0} = -1\). So, the slope of the tangent line at the point (1, 1) is -1.
03

Form the Equation of the Tangent Line

Once the slope of the tangent line is found, use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point and \(m\) is the slope. Substituting the given point (1, 1) and the slope -1 in the equation, we get the equation of the tangent line as \(y - 1 = -1(x - 1)\). Simplifying the equation gives \(y = 2 - x\).

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Key Concepts

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

Derivative
Derivatives are fundamental in calculus and measure how a function changes as its input changes. Simply put, the derivative of a function is the function's instantaneous rate of change or the slope of the function's graph at a particular point. In many cases, derivatives allow us to understand physical concepts such as speed, acceleration, or other rates of change.
  • The derivative of a constant is zero since a constant does not change.
  • The derivative of a function like \( e^x \) is \( e^x \), maintaining its form after differentiation.
For composite functions, such as \( f(x) = e^{1-x} \), the chain rule is often needed to find the derivative.
Tangent line
A tangent line to a curve at a given point is a straight line that "just touches" the curve at that point. It has the same slope as the curve does at that exact point. To find the equation of a tangent line, you need two pieces of information:
  • The slope at the point of tangency, which is obtained from the derivative.
  • The coordinates of the given point on the curve.
Once you have the slope, you can use the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) are the coordinates of the given point. In our example, with \( m = -1 \) and point \( (1, 1) \), this results in \( y = 2 - x \).
Chain rule
The chain rule is a technique in calculus used to differentiate composite functions. A composite function is a function made by combining two or more functions. The chain rule essentially says, to find the derivative of a composite function, you:
  • Differentiate the outer function while keeping the inner function unchanged.
  • Multiply this by the derivative of the inner function.
Using function \( f(x) = e^{1-x} \) as an example:The outer function is \( e^u \), and the inner function is \( u = 1-x \). First, find the derivative of the outer function, \( d(e^u)/du = e^u \). Then, find the derivative of the inner function, \( du/dx = -1 \). Combine these to get the overall derivative \( f'(x) = e^{1-x}(-1) = -e^{1-x} \). The chain rule helps simplify and accurately find derivatives of complex functions.

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Most popular questions from this chapter

Linear and Quadratic Approximations Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a .\) Use a graphing utility to graph the function and its linear and quadratic approximations. $$ f(x)=\tanh x, \quad a=0 $$

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Prove that \(\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad-1

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Is the converse of the second part of Theorem \(5.7\) true? That is, if a function is one-to-one (and therefore has an inverse function), then must the function be strictly monotonic? If so, prove it. If not, give a counterexample.

In Exercises 81 and 82, find \(d y / d x\) at the given point for the equation. \(x=y^{3}-7 y^{2}+2, \quad(-4,1)\)
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