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Chapter 7: Problem 45

Find an equation of the tangent line to the graph of \(f\) at the point \((2, f(2)) .\) Use a graphing utility to check your result by graphing the original function and the tangent line in the same viewing window. $$ f(x)=\sqrt{x^{2}-2 x+1} $$

Short Answer

Expert verified
The equation of the tangent line to the given function at the point (2, 1) is \(y = x - 1\).

Step by step solution

01

Define the function

Firstly, define the function \(f(x)=\sqrt{x^{2}-2x+1}\).
02

Derive the function

The next step is finding the derivative of this function \((f'(x))\). For that, rewrite the function as \(f(x)=(x^2 - 2x + 1)^{0.5}\), and then apply the chain rule for differentiation: \(f'(x) = 0.5*(x^2 - 2x + 1)^{-0.5} * (2x - 2)\). Simplify to obtain \(f'(x) = (x-1)/\sqrt{x^2 - 2x + 1}\).
03

Evaluate the derivative

Evaluate the derivative \(f'(x)\) at \(x=2\) to find the slope of the tangent line: \(f'(2) = (2-1)/\sqrt{2^2 - 2*2 + 1} = 1\).
04

Use point-slope form to find the equation of the line

The point-slope form of the line is \(y - y1 = m(x - x1)\), where m is the slope of the line and \((x1, y1)\) is a point on the line. We know that the slope \(m = f'(2) = 1\) and the point is \((2, f(2)) = (2, 1)\). Substituting these values into the equation, we get \(y - 1 = 1 * (x - 2)\). Upon simplifying, we get the equation of the tangent line as \(y = x - 1\).

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Most popular questions from this chapter

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