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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 7, Problem 7.R.31

Linear approximation Find the linear approximation to ƒ(x) = cosh x at a = ln 3 and then use it to approximate the value of cosh 1.

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Recall that the linear approximation of a function ƒ(x) at a point a is given by the formula: \[L(x) = ƒ(a) + ƒ'(a)(x - a)\] where ƒ'(a) is the derivative of ƒ evaluated at x = a.
Identify the function and the point of approximation: here, ƒ(x) = \(\cosh\) x and a = \(\ln\) 3.
Calculate ƒ(a) by evaluating \(\cosh\)(\(\ln\) 3). Remember that \(\cosh\) x = \(\frac{e^x + e^{-x}\)}{2}, so substitute x = \(\ln\) 3 to find ƒ(a).
Find the derivative of ƒ(x). Since ƒ(x) = \(\cosh\) x, its derivative is ƒ'(x) = \(\sinh\) x. Then evaluate ƒ'(a) = \(\sinh\)(\(\ln\) 3). Use the definition \(\sinh\) x = \(\frac{e^x - e^{-x}\)}{2} to compute this.
Write the linear approximation formula using the values found: \[L(x) = \cosh(\ln 3) + \sinh(\ln 3)(x - \ln 3)\] Finally, substitute x = 1 into this linear approximation to estimate \(\cosh\) 1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Linear Approximation

Linear approximation uses the tangent line at a point a to estimate the value of a function near a. It is given by L(x) = f(a) + f'(a)(x - a), providing a simple way to approximate function values close to a.
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Derivative of cosh x

The derivative of cosh x is sinh x, which represents the slope of the tangent line to the curve at any point x. Knowing this derivative is essential to find the slope f'(a) needed for the linear approximation.
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Evaluating at a specific point

After finding the linear approximation at a = ln 3, you substitute x = 1 into the linear function to approximate cosh 1. This step applies the approximation formula to estimate values near the chosen point.
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