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.
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.
Derivative of the Natural Exponential Function (e^x)
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.