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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 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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