Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.2 Exercises: 81


$y = \frac{1}{4}x + 1$

Work Step by Step

In this exercise, we are given a point $(x_{0}, y_{0})$ that is in a line tangent to $f(x)$, but that is not part of the function itself. In other words, $(x_{0}, y_{0})$ is not the point of tangency between $f(x)$ and the line. As such, the approach is somewhat different to previous exercises. We can construct the following equation for the tangential line using the $(x_{0}, y_{0}) = (-4, 0)$ provided: $$y - y_{0} = m(x - x_{0})$$ $$y - 0 = m(x -(-4))$$ $$y = m(x +4)$$ where $m$ is the slope of the tangential line. We know that, at the point of tangency, both the function $f(x)$ and the equation above must have the same $(x, y)$. Therefore: $$f(x) = \sqrt x$$ $$\sqrt x = m(x +4)$$ for the point of tangency, where by rewriting the equation, we get: $$\frac{\sqrt x}{x+4} = m$$ We can now proceed to find the first derivative of the original function, since we know it gives us a formula for finding the slope of a tangential line to any function $f$: $$f(x) = \sqrt x$$ $$f(x) = x^{\frac{1}{2}}$$ $$f'(x) = \frac{1}{2}x^{\frac{1}{2} - 1}$$ $$f'(x) = \frac{1}{2}x^{-\frac{1}{2}}$$ $$f'(x) = \frac{1}{2\sqrt x}$$ Since we know that $f'(x) = m$, we can now proceed to do the following: $$\frac{1}{2\sqrt x} = \frac{\sqrt x}{x+4}$$ and solve for $x$: $$(1)(x + 4) = (2\sqrt x)(\sqrt x)$$ $$x + 4 = 2x$$ $$x = 4$$ This is the $x$ at the point of tangency, which means that the slope for the tangent line is: $$f'(4) = \frac{1}{2\sqrt 4}$$ $$f'(4) = \frac{1}{4}$$ Therefore, using the first equation we constructed for the tangential line $y = m(x+4)$, we finally get $$y = \frac{1}{4}(x + 4)$$ $$y = \frac{1}{4}x + 1$$
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